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H: Prove a set is numerable I have 4 different sets: a) $\{ [x]: x \in \mathbb{R}\}$ b) $\mathbb Q \cup (2,+\infty)$ c) $\{n^2: n \in \mathbb N\}$ d) $\mathbb Q \cap (2,+\infty)$ A set is numerable if exists a bijective function from $\mathbb N$ to $A$ the c) is numerable because exists $f:\mathbb N \rightarrow A$ so that $f(n)=n^2$ but in the other cases? AI: $a)$ is numerable because the set is identical with $\mathbb Z$ $b)$ is not numerable : Already the interval $(2,3)$ is uncountable. $d)$ is numerable because it is an infinite subset of $\mathbb Q$
H: we have 1000 copies of Hunter's, Rosen's, Liu's and Epp's discrete mathematics books and 1000 students . We have 1000 copies of Hunter's,1000 copies of Rosen's, 1000 of Liu's and 1000 copies of Epp's discrete mathematics books and 1000 distinct students . Every student should take exactly one copy of some discrete mathematics book. If there should be distributed at least one copy of every book , in how many ways can this be done? My teacher's answer uses the "Inclusion-Exclusion" principle. We can violate the constraint if only one category is not distributed at all by $C(4,1)3^{1000}$ ways, or if 2 categories are not distributed at all by $C(4,2)2^{1000}$ or if 3 are not distributed at all by $C(4,3)1^{1000}$ ways . In total : $4^{1000}-C(4,1)3^{1000} + C(4,2)2^{1000} - C(4,3)1^{1000}$ My solution was this : We firstly need to make sure that at least 1 copy of every category is distributed so we pick 4 students with $C(1000,4)$ ways and we can give them book's with $C(1000,4)4!$ ways . Then we are left of with 996 students. Those 996, can pick a book with $4^{996}$ ways . In total : $C(1000,4)4!*4^{996}$ . Can you help me find my bug ? #update # my initial solution is totally wrong , amazing feedback follows. Εxtra question : Any tips to realise this needs Inclusion - Exclusion proinciple? AI: You are considering all the different ways 4 people can be chosen and distributing 4 of the books to them and randomly assigning other books to the others. the problem is that you are counting one case multiple times in different calculations. Assume in scenario P: In your initial choice of 4 people, you choose people 1, 2, 3, and 4 and assign them books A, B, C, and D respectively in one of your calculations. now you assign book A to person number 5, B to 6, C to 7, D to 8, then again A to 9, and so on. Now in scenario Q: In your initial choice of 4 people, you choose people 5, 6, 7, and 8 and assign them books A, B, C, and D respectively in one of your calculations. now you assign book A to person number 1, B to 2, C to 3, D to 4, then again A to 9, and so on. You are counting cases P and Q (and actually many other cases that are the same) repeatedly while they represent the same assignment of books. This is why using the inclusion-exclusion principle is the best way to solve this kind of problem without getting confused. The inclusion-exclusion principle prevents you from calculating the same assignment in different parts of your calculation.
H: Why are we always assuming normality? For a great number of statistical testing (ANOVA or discriminant analysis for example), we suppose that the variables follow a normal law. How many times did I see this supposition ? But I recently learnt that we could calculate the Kurtosis (the measure of the "tailedness") and the Skewness of the variables and check if they both followed the normal value (1 for the kurtosis, and 0 for the skewness). But before that, I always made the assumption that my variables followed a normal law, without testing it. Is this because I am a student, and I just have to apply the theorem and calculus I learn during my scholartime, or is it just because we always assume our variables follow normal law? Furthermore, one of my professor just told me that with a great amount of data (~ >= 1 million), we just assume that the variables and its moment estimators follow a normal law. But when I calculated my kurtosis and skewness, I found enormous values, sometimes greater than 100. To me, it means my variables DOESN'T follow a normal law. So why my professor assume it does? I apology for my poor English, and I hope my question is clear nonetheless. AI: There are multiple questions that you ask. In general, assuming normality is plain wrong, a random variable could be whatever (exponential, Pareto, beta, gamma). In the same sense if you have some data it could be that they come from a particular distribution or that they are "random" or deterministic. About the skewness and kurtosis, you can calculate them and see if they match those of a normal distribution, be aware this matching does not mean that your data come from a normal distribution necessarily. Finally, any amount of data does not allow you to assume normality (why would it?), most probably your professor was talking about the Central limit theorem which says something totally different. On a high level the Central limit theorem says : If you have a population with mean $\mu$ and standard deviation $\sigma$ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed.
H: Closedness of a subset of complex numbers under addition If we have two angles $$\phi_1,\phi_2\in[0,2\pi]$$ such that $$\phi_1\le\phi_2$$ and we perform standard addition on complex numbers from the subset of $$S=\{ z\in \mathbb{C} : arg(z) \in[\phi_1 ,\phi_2]\}$$ do we get a closed algebraic structure $(S,+)$? In other words, for every $z_1,z_2 \in S$, is $z_1+z_2 \in S$ ? Intuitively it makes sense to me that this should be closed, since if we add 2D geometrical vectors (represented by the complex numbers) whose arguments are bounded, using the geometrical rules for vector addition (eg. the parallelogram rule) the resultant vector (complex number) has a bounded argument to the same boundaries as the initial two vectors. I can't find any counterexamples, but I also cannot figure out how to show closedness. I've expressed the arguments of $z_1=x_1+iy_1$ and$z_2=x_2+iy_2$ as $\theta_1 = \arctan{(y_1/x_1)}$ and $\theta_2 = \arctan{(y_2/x_2})$, but I don't see how to express the argument of their sum $arg(z_1+z_2)=\theta_{12} = \tan(\frac{y_1+y_2}{x_1+x_2})$ in terms of their arguments or how to show that $\theta_{12}\in[\phi_1 ,\phi_2]$. I know that I can use the identity for the tan of an angle sum (which we would have if we were multiplying the complex numbers $z_1$ and $z_2$) to express it in terms of the tans of the $z_1$ and $z_2$, but I don't see that as being the case here. EDIT: Note to self - when looking for counter-examples, look more at the extremes. AI: If $\phi_2-\phi_1<\pi$ you get a sector in the complex plane between two half-lines. That's closed under addition. When $\phi_2-\phi_1=\pi$ you get a half-plane, but when $\phi_1-\phi_0>\pi$ you get a sector with a reflex angle, and that's not closed under addition. As a concrete example, take $\phi_1=0$, $\phi_2=3\pi/2$, $z_1=1$ and $z_2=-i$. Then $z_1+z_2=1-i\notin S$.
H: Markov Property definition via conditional expectation In several textbooks I have seen the following equivalent statement for the Markov property: Let $\{X\}_{t \geq 0}$ be a stochastic process, $\mathcal{F}_u^v = \sigma\{X_s, s \in [u,v]\}$. Then $\{X_s\}_s$ has the Markov property iff for all $0 \leq t < s$ and any any bounded Borel function $g$ it is $$ \mathbf{E}\{g(X_s)\|\mathcal{F}_0^t\} = \mathbf{E}\{g(X_s)\| \mathcal{F}_t^t\}.$$ I wonder whether the condition of $g$ being bounded is really necessary. Does the statement hold for any Borel function $g$ with $\mathbf{E}\|g(X_s)\| < \infty$ as well? AI: Yes. If the equation holds for bounded measurable functions it holds for simple functions. By Monotone convergence Theorem it holds for all non-negative measurable functions. Hence it holds whenever $E\|g(X_s)\| <\infty$.
H: Two natural numbers are selected at random, what is the probability that the sum is divisible by 11? This same question to check divisibility by 10 is quite straightforward but when we have to check whether it is divisible by 11 there is one case that has to be excluded, I would like it if someone explains it intuitively. edit: Nvm its not very different from 10 or any other number n AI: If you have fixed one number, the other can have $11$ possible residues mod $11$, one of which is suitable. Hence the answer is simply $\frac{1}{11}$
H: $\frac{dy}{dx}=\frac{y^3}{e^x +y^2}$ The curve $f(x,y)=0$ passes through $(0,2)$ and satisfies $$\frac{dy}{dx}=\frac{y^3}{e^x +y^2}$$ The line $x=\ln 5$ intersects the curve at $y=a$ and $y=b$. Find the value of $\frac{4(a^2+b^2)}{53}$ It is the differential equation that I'm unable to crack. I'm almost sure that it is reducible to a Linear Differential Equation. My futile attempts include: Homogeneous form/reducible to homogeneous form Linear Differential treating $x$ as independent variable I am also toying with the idea that here $y$ could be the independent variable but I'm unable to proceed. Any help would be most welcome. AI: Hint $(1/y^2+e^{-x})dy=e^{-x}ydx \implies dy/y^2 = -d(e^{-x}y)$
H: $\sum_{n=1}^\infty\frac{(-1)^n\sin ^2n}{n}$ Is the following solution wrong ?; Does $\sum\frac{(-1)^n\cos 2n}{2n}$ converge? $$\sum_{n=1}^\infty\frac{(-1)^n\sin ^2n}{n}$$ Solution from the lecture notes : $$\frac{(-1)^n\sin ^2n}{n}=\frac{(-1)^n(1-\cos > 2n)}{2n}=\frac{(-1)^n}{2n}-\frac{(-1)^n\cos 2n}{2n}$$ $\frac{(-1)^n}{2n}$ converges conditionally due to Leibniz test. $\frac{(-1)^n\cos 2n}{2n}$ converges absolutely since $\frac{\cos 2n}{2n}$ converges due to Dirichlet test. We conclude that the sum we started with coverges conditionally. $\frac{(-1)^n\cos 2n}{2n}$ cannot converge absolutely because $$\left \|\frac{(-1)^n\cos 2n}{2n} \right \|=\left \|\frac{\cos 2n}{2n} \right \|>\frac{\cos^2 2n}{2n}=\frac{\frac12(1+\cos4n)}{2n}=\frac1{4n}+\frac{\cos4n}{4n}$$ This is a sum of a convergent and divergent series hence the series doesn't absolutely converge. Am I correct and does $\sum\frac{(-1)^n\cos 2n}{2n}$ converge ? AI: Dirichlet test with $b_k = \frac{1}{k}$ monotone decreasing and converging to $0$ and $a_k = (-1)^k \sin^2 k$: $$ S_M = \|\sum_{k=1}^{M}(-1)^k \sin^2(k) \|\leq \|\sum_{k=1}^{M}(-1)^k \| \leq 1\\ $$
H: Show that $ x\cdot\cos(x)+\sin(x)/2=\sum_{n=2}^\infty (-1)^n\cdot\frac{2n}{n^2-1}\cdot\sin(nx)$ when $x\in [-\pi,\pi]$ To show this, I have used definitions for $\cos(x)$ and $\sin(x)$: $$x\cdot \cos(x)+1/2\sin(x)=x\cdot \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}\cdot x^{2n}+1/2\cdot \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\cdot x^{2n+1}$$ However, i do not how I'm supposed to proceed from here. You've got any ideas? This comes from the fourier series of $x\cdot \cos(x)$. We have the fourier coefficients as: $$ c_{-1}=-\frac{i}4, c_0=0,c_1=\frac{i}4$$ and $$c_n=(-1)^{n-1} \cdot \frac{in}{n^2-1}$$ when $\|n\|\geq 2$ AI: Hint: $$\dfrac{2n\sin nx}{(n+1)(n-1)}=\dfrac{\sin nx}{n-1}-\dfrac{\sin nx}{n+1}$$ Now $(-1)^n\dfrac{\sin nx}{n-1}$ is the imaginary part of $$\dfrac{(-1)^ne^{inx}}{n-1}=e^{ix}\cdot-\dfrac{(-e^{ix})^{n-1}}{n-1}$$ Now $$\sum_{n=2}^\infty-\dfrac{(-e^{ix})^{n-1}}{n-1}=\ln(1+e^{ix})=\ln (e^{ix/2})+\ln\left(2\cos\dfrac x2\right)=(2n\pi+\dfrac x2)i+\ln\left(2\cos\dfrac x2\right)$$ Put $n=0$ to find the principal value Similarly for $$\dfrac{\sin nx}{n+1}$$ Finally use How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?
H: How to show $x_{k}=\frac{(I-\alpha Q)^{k} x_{0}}{\left\\|(I-\alpha Q)^{k} x_{0}\right\\|}$ where Q is positive definite, converges to optimal solution I'm trying to show $\overrightarrow x_{k}=\frac{(I-\alpha Q)^{k} \overrightarrow x_{0}}{\left\\|(I-\alpha Q)^{k} \overrightarrow x_{0}\right\\|}$ where $\overrightarrow x_{k}$ converges to an optimal solution as k tends to infinity. This is an update equation in Optimization. Q is positive definite, real nxn, and the eigenvalues are distinct. $ x_{0}$ is also not orthogonal to the eigenvector of Q with smallest eigenvalue I need to show for $0<\alpha<1 / \lambda_{\max }$ , the fixed step size converges to an optimal solution. $\lambda_{\max }$ max eigenvalue of Q My attempt is to try and factorize. so $x_{k}=\frac{(I-\alpha V\Sigma V^{T})^{k} \sum ^{n}_{i=1}\mu _{i}\overrightarrow {v_{i}}}{\left\\|I-\alpha V\Sigma V^{T})^{k} \sum ^{n}_{i=1}\mu _{i}\overrightarrow {v_{i}}\right\\|}$ and take $\lim _{k\rightarrow \infty }$ Where $v_{i}$ is eigenbasis of Q And then something with the terms to the power of k tending to 0. AI: This might help: $(I-\alpha Q)=V(I- \alpha D)V^T$ where $V$ is orthogonal and $D$ diagonal and $VDV^T=Q$. $(I-\alpha Q)^k=V(I- \alpha D)^k V^T$ note that $(I- \alpha D)$ is diagonal, so $(I- \alpha D)^k$ are just the diagonal entries to the power of $k$. Denote $\lambda_i=D_{ii}$ the $i$-th eigenvalue. For the $i$-th diagonal entry it holds: $(I- \alpha D)^k_{ii}=(1-\lambda_{min}\alpha)^k \frac{(1-\alpha \lambda_i)^k}{(1-\lambda_{min}\alpha)^k}$ $\frac{(I- \alpha D)^k_{ii}}{(1-\lambda_{min}\alpha)^k}= \frac{(1-\alpha \lambda_i)^k}{(1-\lambda_{min}\alpha)^k}$ In the limit, the right hand side is equal $1$ if $\lambda_i=\lambda_{min}$ and $0$ else. This implies $\frac{V(I- \alpha D)^kV^T}{(1-\lambda_{min}\alpha)^k}\to v_{min}v_{min}^T$ Now, it is easy to show, that $x_k \to v_{min}$ (the eigenvector to the smalest eigenvalue) Edit: For the formal proof, use $ x_{k}=\frac{\frac{(I-\alpha Q)^{k} }{(1-\lambda_{min}\alpha)^k} x_{0}}{\left\\|\frac{(I-\alpha Q)^{k}}{(1-\lambda_{min}\alpha)^k} x_{0}\right\\|}\to \frac{v_{\min}v_{\min}^T x_0}{\left\\|v_{\min}v_{\min}^T x_0\right\\|}=v_{\min}$ We can take these limits because everything is finite.
H: Why when $t\equiv n\pmod s$, $n$ is also $t - s\lfloor \frac{t}{s}\rfloor$? I'd like to know why when $t\equiv n\pmod s$, $n$ is also $t - s\lfloor \frac{t}{s}\rfloor$? Here, $n\in[0;s[$ I can't find a way to prove $n$. Thank you for your help! AI: $n$ is the remainder when $t$ is divided by $s$, which you can write as $t = n+sq$. Since the quotient, $q$, is the integer part of the result of the division, $q = \left\lfloor\frac{t}{s}\right\rfloor$. Substitute that in and do a basic transposition, and you get $n = t-s\left\lfloor\frac{t}{s}\right\rfloor$.
H: Is $( \mathbb{ Z}^,\cdot) \rightarrow (\mathbb{Z}_5^{},\cdot), n \mapsto n \pmod 5 $ well-defined? Is $( \mathbb{ Z}^,\cdot) \rightarrow (\mathbb{Z}_5^{},\cdot), n \mapsto n \pmod 5 $ well-defined? So what I think is that it is not well -defined because the non-zero multiples of 5 in $\mathbb{ Z}^$ map to $[0]_5$ but $[0]_5 \notin \mathbb{Z_5^}$ What do you think? AI: You seem to think that $\mathbb{Z}^=\mathbb{Z}\backslash\{0\}$. That would be a rather uncommon notation and it is likely to be wrong. At least in pure algebra. Given a unital ring $R$, typically $R^$ denotes the group of all invertible elements of $R$: $$R^=\big\{x\in R\ \big\|\ xy=yx=1\text{ for some }y\in R\big\}$$ which is a group under multiplication inherited from $R$. Note that $0\not\in R^$ unless $R$ is a trivial (one element) ring. And so $R^\subseteq R\backslash\{0\}$, but $R^$ may and often is a proper subset of $R\backslash\{0\}$. In fact $R^=R\backslash\{0\}$ if and only if $R$ is a division ring (or field in commutative case). So for example $\mathbb{Z}_5^$ is equal to $\mathbb{Z}_5\backslash\{0\}$, but only because $\mathbb{Z}_5$ is a field. But $\mathbb{Z}$ is not a division ring and in fact $\mathbb{Z}^=\{-1,1\}$. It is literally the set containing two elements: $-1$ and $1$, no other integer is invertible in $\mathbb{Z}$. So multiplicies of $5$ are not even in $\mathbb{Z}^$. You can easily verify that with this $n\mapsto n\text{ mod }5$ is a well defined group homomorphism. In fact if $f:R\to S$ is any unital ring homomorphism, then it induces a group homomorphism $f^:R^\to S^$ given by $f^(x)=f(x)$. We can apply this to the well known unital ring homomorphism $\mathbb{Z}\to\mathbb{Z}_m$, $n\mapsto n\text{ mod }m$. Edit: Unless you actually do define $\mathbb{Z}^*$ as $\mathbb{Z}\backslash\{0\}$. In that case your reasoning is correct and this is not a well defined function.
H: Defining the logarithm of a function I have a contractible open subset $U$ of a smooth manifold $M$, and a smooth $f:U\to\mathbb C^$ (I don't know if I need all of these properties, but it is what I am working with). Can I define $\log(f):U\to\mathbb C$, such that $\exp\circ\log(f)=f$? AI: Yes you can define $\log(f)$ and it suffices that your open subset $U$ is simply connected and $f$ is continuous. What you are looking for is a lift $g$ of $f$ to the covering $\exp : \mathbb{C} \rightarrow \mathbb{C}^$; it follows from the lifting property of coverings that such a lift exists provided that $f_{\#}(\pi_1(U,u))$ is trivial, which is always satisfied if $U$ is simply connected. You also have that such a lifting is unique once you fix the image $g(u)$ of $u \in U$.
H: Why is a solution colinear to two rational solutions of a Weierstrass-form elliptic curve also rational? I learned this fact and it blew my mind: given an equation $y^{2}=x^{3}+ax+b$ and two rational solutions: $(x_1, y_1), (x_2, y_2)$ with $x_1, y_1, x_2, y_2 \in \mathbb{Q}$, then any other solution colinear with the first two solutions is also rational; that is to say, any solution to the system: $y^{2}=x^{3}+ax+b \\ y=y_1+\frac{y_2-y_1}{x_2-x_1}(x-x_1)$ has the property: $x, y \in \mathbb{Q}$ How would you go about proving this? Could you perhaps sketch out the main steps of the proof? Also, if you could, please share any resources / tutorials / textbooks that could help me learn the necessary mathematics for this proof; I've already looked at a few algebraic geometry textbooks and I found it quite difficult; my background is statistics and machine learning (PhD), but I find reading pure algebra just incredibly difficult due to the sheer number of definitions they can have on a single page and still use a large number of terms I don't know. AI: You don't need any advanced math for this. What you are saying is that if a degree 3 polynomial $f=ax^3+bx^2+cx+d\in \mathbb Q[x]$ has 2 roots in $\mathbb Q$, then all its roots are in $\mathbb Q$. In fact, you can alway write $f=a(x-x_1)(x-x_2)(x-x_3)$, where $a\in \mathbb Q^*$ and $x_1,x_2,x_3$ are its roots. Now assume wlog $x_1,x_2\in \mathbb Q$. Then $-a(x_1+x_2+x_3)=b\in\mathbb Q$. But $a,x_1,x_2\in \mathbb Q$ and therefore $x_3\in \mathbb Q$ as well. If you're interested in the topic and you want a soft introduction, I recommend Silverman-Tate's book "Rational points on elliptic curves".
H: Expected value of a sum of die rolls You roll a standard six-sided die, then roll n more six-sided die, where n was the first roll. What is the expected value of the sum of all the die you rolled? I did some calculations, but I'm not sure. Let expected value of the sum of all the die rolls be E[X], then $$E[X] = \frac{3.5}{6}+2\frac{3.5}{6}+3\frac{3.5}{6}+4\frac{3.5}{6}+5\frac{3.5}{6}+6\frac{3.5}{6}$$ because the expected value of each single die is 3.5. Is it correct? AI: Let $N$ be the outcome of the first six-sided die. Then you want to find $$ \mathbb{E}[N + \sum_{i = 1}^N X_i] $$ where $X_i$ is the outcome of the $i$-th six-sided die. The trick is to condition on the value of $N$. Assuming that $N$ can be any value between 1,..,6 we have: $$ \mathbb{E}[N + \sum_{i = 1}^N X_i] = \sum_{j = 1}^6 \mathbb{E}[N + \sum_{i = 1}^N X_i \| N = j]\mathbb{P}(N = j) = \sum_{j = 1}^6 (j + \mathbb{E}[\sum_{i = 1}^j X_i])\mathbb{P}(N = j), $$ Then you can fill in the known $\mathbb{P}(X = j)$ and use that $$ \mathbb{E}[\sum_{i = 1}^j X_i] = \sum_{i = 1}^j \mathbb{E}[X_i] $$ when $i,j$ are known.
H: Prove $2\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/2\mathbb{Z} \overset{\sim}{=} \mathbb{Z}/2\mathbb{Z}$ I need to prove that $2\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/2\mathbb{Z} \overset{\sim}{=} \mathbb{Z}/2\mathbb{Z}$. I know that $\mathbb{Z} \otimes_{\mathbb{Z}}\mathbb{Z}/2\mathbb{Z} \overset{\sim}{=} \mathbb{Z}/2\mathbb{Z}$, but using the universal property of the tensor prodcut doesn't seem to work here. Advice is greatly appreciated. AI: Hint: As a $\mathbf Z$-module, $\;2\mathbf Z\simeq\mathbf Z$. Hence…
H: What is $\sin{ω}$? I am reading about hyperreal numbers defined as (to my understanding) certain equivalence classes on all sequences of real numbers. $ω$ is defined as $(1, 2, 3, ...)$, and all functions are applied element-wise. This makes sense for sequences that have an infinite limit, like $e^ω$, which is simply a bigger infinity than $ω$, but it occurs to me that there must be certain "abominable" numbers that don't fit into the standard notion of reals, omegas, and epsilons. By the transfer principle, values like $\sin{ω} = (\sin{1},\sin{2},\sin{3},...)$ and $\frac{\sin{ω}}{ω}=(\frac{\sin{1}}{1},\frac{\sin{2}}{2},\frac{\sin{3}}{3},...)$ should exist and be finite/limited. However, for the first value I am not able to show that it is equal to a real number, or at least infinitely close to a real number (which is a necessity according to the paper I am reading), and for the second number, although it is infinitely close to $0$, its sign is ambiguous, breaking trichotomy (in the same way as $\sin{ω}$) which should hold based on the transfer principle. So, if $\sin{ω} = r$, what is the actual value of $r$? AI: Your questions show that you have yet to internalize the following important fact: your source constructs the hyperreals $\:^\star\mathbb{R}$ in Section 1.3. by first fixing a non-principal ultrafilter $\mathcal{F}$ over $\mathbb{N}$. You use $\omega$ to denote the fixed hyperreal arising from (the equivalence class of) the sequence $(0,1,2,\dots)$. But the properties of fixed hyperreals such as $\omega$ actually depend on the precise identity of the ultrafilter $\mathcal{F}$ you used to construct $\:^\star\mathbb{R}$. Before we look at trigonometric functions, you should first think about something simpler, say the function $f: \mathbb{R} \rightarrow \mathbb{N}$ given as follows: $$f(x) = 1 \text{ if }x\text{ is an odd integer and }f(x)=0\text{ otherwise.}$$ By Section 1.6. of your source, we can get an extended function $^\star\!f: \:^\star\mathbb{R} \rightarrow \:^\star\mathbb{N}$. But what is the value of $^\star\!f(\omega)$? Since $\forall x \in \mathbb{R}. f(x) = 0 \vee f(x) = 1$, the Transfer principle says that either $^\star\!f(\omega) = 0$ or $^\star\!f(\omega) = 1$. But which? It turns out that the answer depends on the ultrafilter $\mathcal{F}$ you used to construct $\:^\star\mathbb{R}$. You should check that if the ultrafilter contains the set of odd natural numbers, then $^\star\!f(\omega) = 1$, and if instead the ultrafilter contains the set of even natural numbers, then $^\star\!f(\omega) = 0$. Since for any set $A \subseteq \mathbb{N}$ $\mathcal{F}$ contains either $A$ or $\mathbb{N}\setminus A$, it must be one of these. Once you understand these results, we can move on to your questions about the $\sin$ function. However, for the first value I am not able to show that it is equal to a real number, or at least infinitely close to a real number (which is a necessity according to the paper I am reading Of course $\sin(\omega)$ is infinitesimally close to some real number. This follows immediately from the general result that every limited hyperreal has a shadow (Theorem 3.3. in your source). And of course $\sin(\omega)$ is limited, since we have $\forall x. \|\sin(x)\| < 2$. But which real number $r$ is $\sin(\omega)$ infinitesimally close to? Well, that depends strongly on the non-principal ultrafilter $\mathcal{F}$. In fact, since the image of $\sin(\mathbb{N})$ is dense in the interval $[-1,1]$, you can choose any number $x \in [-1,1]$ and find an ultrafilter $\mathcal{F}$ such that $\sin(\omega) \approx x$ in the hyperreal field constructed using $\mathcal{F}$. and for the second number, although it is infinitely close to 0, its sign is ambiguous, breaking trichotomy For similar reasons, the value of $\mathrm{sgn}\left(\frac{\sin(\omega)}{\omega}\right)$ is very far from ambiguous: it is either positive (so the $\mathrm{sgn}$ function takes the value $1$ or negative ($-1$), and not both. To deduce which possibility holds, you have to know more about the non-principal ultrafilter that was used to construct $\mathcal{F}$: if $\mathcal{F}$ contains the set $\left\{n \in \mathbb{N} \:\|\: \frac{sin(n)}{n} > 0\right\} \subseteq \mathbb{N}$ then $\frac{\sin(\omega)}{\omega} > 0$ (exercise!) so its sign is positive. If $\mathcal{F}$ contains the complement of this set, then the sign is negative (exercise: why can't it be zero?). And $\mathcal{F}$ must contain one of these, by virtue of being an ultrafilter.
H: Which of the following is divergent? $\sum\frac1n\sin^2\frac1n$, $\sum\frac1{n^2}\sin\frac1n$, $\sum\frac1n\log n$, $\sum\frac1n\tan\frac1n$ Which of the following is divergent? (a) $\displaystyle \quad\sum_{n=1}^{\infty}\frac{1}{n}\sin^2(\frac{1}{n})$ (b) $\displaystyle \quad\sum_{n=1}^{\infty}\frac{1}{n^2}\sin(\frac{1}{n})$ (c) $\displaystyle \quad\sum_{n=1}^{\infty}\frac{1}{n}\log n$ (d) $\displaystyle \quad\sum_{n=1}^{\infty}\frac{1}{n}\tan (\frac{1}{n})$ My attempt: $\sin(x)\le x$ so for option (b) $\sum_{n=1}^{\infty}\frac{1}{n^2}\sin(\frac{1}{n})\le\sum_{n=1}^{\infty}\frac{1}{n^2}$ which is convergent. Hence, series in (b) is convergent. For option (a) $\sum_{n=1}^{\infty}\frac{1}{n}\sin^2(\frac{1}{n})\le\sum_{n=1}^{\infty}\frac{1}{n^3}$ . However, I am not sure if the inequality I have used is correct. Can someone please check this and explain how to workout for options (c) and (d). Thanks for the help ! AI: You have$$\lim_{n\to\infty}\frac{\frac1n\sin^2\left(\frac1n\right)}{\frac1{n^3}}=1$$and therefore the first series converges. You always have (unless $n$ is $1$ or $2$) $\frac{\log n}n>\frac1n$ and therefre the third series diverges. And$$\lim_{n\to\infty}\frac{\frac1n\tan\left(\frac1n\right)}{\frac1{n^2}}=1.$$So, the fourth series converges.
H: Prove that $\int_{\Omega}fd\mu \int_{\Omega}gd\mu\geq 1.$ Suppose that $\mu(\Omega)=1$ and suppose $f$ and $g$ are positive measurable functions on $\Omega$ such that $fg\geq 1$. Prove that $$\int_{\Omega}fd\mu \int_{\Omega}gd\mu\geq 1.$$ AI: $\phi (x)=\frac 1 x$ is a convex function on $(0,\infty)$. By Jensen's inequality we get $\phi (\int fd\mu)=\frac 1 {\int f d\mu} \leq \int \frac 1 f d\mu \leq \int g d\mu$ since $\frac 1 f \leq g$.
H: Is $(\mathbb{Z_4},+) \rightarrow (\mathbb{Z_5^{}},\cdot), n\bmod 4 \mapsto n \bmod 5 $ a homomorphism? For the following relation $(\mathbb{Z_4},+) \rightarrow (\mathbb{Z_5^{}},\cdot), n\bmod 4 \mapsto n \bmod 5 $ Determine if it is well-defined and an homomorphism So I think it is not well-defined because $6\equiv 2 \text{ ( mod 4)}$, but $6\not \equiv 2 \text{ ( mod 5)}$, then it can't be a homomorphism either. What do you think? AI: You're correct. Nevertheless, there exist isomorphisms between these groups, bu they're defined through generators of the group $\mathbf Z^\times$, which are $2$ and $3$, say via $2$: \begin{align} \mathbf Z/4\mathbf Z&\longrightarrow(\mathbf Z/5\mathbf Z)^\times\\ k\bmod4&\longmapsto 2^{k\bmod 4}\bmod 5 \end{align}
H: Limit of Sequence... Suppose $a>0$, define sequence $a_n$ by $$a_n=\sqrt{n}\left(\sqrt[n]{ea}-\sqrt[n]{a}\right).$$ Find $\lim_{n\to \infty} a_n$ if it exists. By observing the speed of $\left(\sqrt[n]{ea}-\sqrt[n]{a}\right)$ I thought that limit is $0$. AI: $a_n=a^{1/n} \sqrt n (e^{1/n}-1)$. Now $a^{1/n} \to 1$ and $e^{1/n}-1 \sim \frac 1 n$ as $n \to \infty$. Hence the limit is $0$.
H: Regarding a result about the degree of an element in a family of finite sets. We have a family $F = \{S_1, S_2... S_m\} $ of $m$ subsets of $\{1,2...n\} $, all with the same cardinality. We're given that $ \forall a,b \in \cup S_i $, number of subsets $S_i$ containing both $a$ and $b$ is the same, or, $\|\{S_i : S_i \in F, a \in S_i, b \in S_i\}\| = l $, for some fixed integer $l$. We need to prove that $ \forall a,b \in \cup S_i $, number of sets containing $a$ is equal to the number of sets containing $b$. (or $degree(a) = degree(b)$.) I attempted to prove it by contradiction, assuming otherwise, let $ \exists a,b \in \cup S_i $ s.t. $deg(a) \neq deg(b)$. Let the family of sets containing $a$ be given by $F_a$. We know that for another $c \in \cup S_i $, $\|F_a\| + \|F_c\| = \|F_{a \cup c}\| + \|F_{a \cap c}\| $. Writing this equation for $b$ as well and subtracting, we get $$ \|F_a\| - \|F_b\| = \|F_{a \cup c}\| - \|F_{b \cup c}\| $$ We need to prove the right side of the expression is zero, but I do not know how to proceed. Any hints? AI: Say number $1$ is in $d$ sets, WLOG we can assume in $S_1,S_2,...S_d$. Also uppose $\|S_i\|=k$ for each $i$ and each pair $\{x,y\}$ appear in $l$ sets among $S_1,...S_m$. Then if we count the connections between set $\{2,3,...,n\}$ and sets $\{S_1,...S_d\}$ we have $$(n-1)l = d(k-1)\implies d = l\cdot{n-1\over k-1}$$ and we are done. We only have to think about the possibility $k=1$. Clarly we can say the same for each vertex $\ne 1$, so $d$ is the same for all vertices.
H: Why is the vector $(u_x, u_y, - 1)$ normal to the surface $u=u(x, y) $? Studying the method of characteristics, the argument goes as follows: We are interested in the equation: $a(x, y)u_x+b(x, y) u_y=f(x, y, u)$; $(a(x, y), b(x, y), f(x, y, u))(u_x, u_y, - 1) =(a,b,f)\nabla{F}=0$, where $F =u(x, y) - u=0$. Hence, $(a, b, f)$ is orthogonal to $\nabla F$. Then, they say that $\nabla F$ is orthogonal to the solution surface, and so we get that $(a, b, f) $ lies in the tangential plane to the solution surface. Therefore, we can build a characteristic curve, starting from a point on the boundary. We parametrise our unknown characteristic curve as ${x(t), y(t), z(t)} $, and find the tangent vector to it at every point - $v=(x_t (t), y_t (t), z_t (t)) $. Then, we find $x, y, z$ from the condition $v=c(a, b, f)$. This is a system of ODEs, comprising the method of Characteristics. However, I don't understand why they say that $\nabla F=(u_x, u_y, - 1)$ is orthogonal to the solution surface. AI: The tangent plane of the surface parametrized by $$\Gamma =\{\sigma (x,y):=(x,y,u(x,y))\mid (x,y)\in I\},$$ at the point $(a,b)$ is given by $$u(a,b)+\text{Span}\Big\{\sigma _x(a,b),\sigma _y(a,b)\Big\},$$(whenever $\{\sigma _x(a,b),\sigma _y(a,b)\}$ is free) and thus, has normal vector $$\sigma _x(a,b)\times \sigma _y(a,b),$$ which is collinear to $(u_x,u_y,-1)$.
H: Every set $A\subseteq\mathbb{R}\;$ is isomorphic to some subset of $[a,b]\;$ for all $b>a$ I want to prove a more general theorem. I know that for some ordinal $\alpha$ there exists an isomorphism between $\alpha$ and some $A\subseteq\mathbb{R}\;$ I want to show that for all $b>a\;$ I can find a subset of $[a,b]\;$ such that $\alpha$ is isomorphic to that subset. AI: It is an immediate consequence of having a strictly increasing embedding $f:(\Bbb R,\le)\hookrightarrow ([a,b],\le)$ in, say, $f(x)=a+\frac{b-a}\pi\operatorname{arccot}(-x)$
H: Relation between $(km)!$ and few others Show that $(m!)^k, (k!)^m, k!m!$ are all less equal $(km)!$ where $k$ and $m$ are integers greater than $1$. I tried it by inducting on $k$ keeping $m$ fixed but got stuck at inductive step. AI: For any $0\leq j\leq k-1$, $$ \prod_{t=1}^m (jm+t)\geq \prod_{t=1}^m t=m! $$ So $$ (km)!=\prod_{j=0}^{k-1}\prod_{t=1}^m(jm+t)\geq \prod_{j=0}^{k-1} m!=(m!)^k $$ Switching the roles of $k$ and $m$ yields $(km)!\geq (k!)^m$. For the last one, assume $k\leq m$ without loss of generality. If $k\geq 2$ then $$ (km)!\geq (m!)^k\geq m!m!\geq k!m!. $$ If $k=1$ then $(km)!=m!=k!m!$. Added later: The first one can be done with a combinatorics argument instead. $(km)!$ is the number of ways of ordering $km$ objects. Split those $km$ objects into $k$ groups of $m$ each. Then the total number of orderings is at least the number of ways of ordering each of the $k$ groups of $m$ individually. The latter figure is $(m!)^k$ since each group has $m!$ orderings and there are $k$ groups.
H: Prove that every random variable has a cumulative distribution function The answer by this post started by stating that every random variable has a cumulative distribution function (CDF). Question: How to prove that the existence of CDF of every random variable? I understand that some random variable might not have probability density function. But I have no idea how to prove the question. AI: It is not a matter of proof but merely of definition. If $X$ is a random variable then we can define the function $\mathbb R\to\mathbb R$ as prescribed by: $$x\mapsto P(X\leq x)$$ Observe that - because $X$ is a random variable - for every $x\in\mathbb R$ the set $\{X\leq x\}$ is a measurable set so that the expression $P(X\leq x)$ makes sense and is an element of $[0,1]\subseteq\mathbb R$. This function is labeled as the CDF of $X$ and is mostly denoted as $F_X$.
H: First order logic: structural induction I am having trouble trying to prove the following. $\mathcal{S}\subset \mathcal{P}(\mathbb{N})$ is the minimal set that satisfies the following: $\mathbb{N} ∈ \mathcal{S} $ for every $a \in \mathbb{N}$, $\mathbb{N} \setminus \{a\} ∈ \mathcal{S}$ for every $A,B \in \mathcal{S}$, $A\cap B \in \mathcal{S}$ prove that $\mathcal{S}$ is the set of all co-finite sets of $\mathbb{N}$ As I understand, this is structural induction, though I don't really know how to approach it. I would appreciate both an explanation of how to use structural induction in general and a solution to this question. Edit: $\mathbb{N}$ is the set of natural numbers AI: $\newcommand{\N}{\mathbb{N}}$ Given a finite set $X \subseteq \N$, we can induct on $\|X\|$ that $\N \setminus X \in S$ (which proves that all cofinite sets lie in $S$). The case is trivial for $\|X\| = 0$, as by hypothesis $\N \in S$. The second condition also proves the case for $\|X\| = 1$. Now suppose for all $X \subseteq \N$ such that $\|X\| = n$, we have $\N \setminus X \in S$. Let $a_1,a_2,\dots,a_n,a_{n+1} \in \N$. Then by inductive hypothesis, we have that: $$ A = \N \setminus \{a_1,a_2,\dots,a_n\} \in S \\ B = \N \setminus \{a_2,\dots,a_n,a_{n+1}\} \in S $$ Therefore, by the third condition: $$ \N \setminus \{a_1,a_2,\dots,a_n,a_{n+1}\} = A \cap B \in S $$ Since $a_1,\dots,a_{n+1}$ are arbitrary, the induction is complete, so all cofinite sets lie in $\N$. Since $S$ is minimal, and the set of all cofinite sets satisfy the three conditions, $S$ must be precisely the set of all cofinite sets.
H: Asymptotic behavior of $\sum_{k=1}^{n^2}\frac{q^{n^2-k}(1-q^k)}{k(1-q^{n^2})}$ when $q=1-\frac{\log(n)}{n}$ Let $q=1-\dfrac{\log(n)}{n}$. Numerical simulations indicate that that \begin{align} \lim_{n \to \infty}\displaystyle\sum_{k=1}^{n^2}\dfrac{q^{n^2-k}(1-q^k)}{k(1-q^{n^2})} = 0 \end{align} in a monotone decreasing manner. But I wonder about the rate of convergence. Any ideas? AI: Your sum is basically $(1-q^{n^2})\sum_{k=0}^{n^2-1}{\frac{q^{k}}{n^2-k}}-\frac{q^{n^2}H_{n^2}}{1-q^{n^2}}$. But $q^{n^2} \leq e^{-n^2 \cdot ((\ln{n})/n)}=n^{-n}$, so the second term is $O(n^{-n}\ln{n})$. For the first term, the same computation shows that $q^{-3n} \leq n^{-3}$, so the sum is less than $\sum_{k=3n}^{n^2-1}{q^k}+(1+O(n^{-n}))\sum_{k=0}^{3n-1}{\frac{q^k}{n^2-k}}+O(n^{-n}\ln{n})$. The first sum is thus at most $O((n^2\ln{n})^{-1})$. The second sum is at most $(n^2-3n)^{-1}\cdot \frac{n}{\ln{n}}$, and therefore the total is $O((n\ln{n})^{-1})$. To make the estimate more precise, we see that the dominant term is the sum $\sum_{k=0}^{3n-1}{\frac{q^k}{n^2-k}}$. Note that $\sum_{k=0}^{3n-1}{\frac{1}{n^2-k}-\frac{1}{n^2}} \leq \frac{1}{n^3(n-3)}\frac{3n(3n-1)}{2} = O(n^{-2})$, so that the first sum is $O(n^{-2})+n^{-2}(1+O(n^n))\sum_{k=0}^{3n-1}{q^k}=O(n^{-2})+\frac{n}{\ln{n}}(1-q^{3n})=\frac{1}{n\ln{n}}+O(n^{-2})$. So your sequence is equivalent to $\frac{1}{n\ln{n}}$.
H: Solving an algebraic equation with fractions How do I rework the following equation to solve for P when I know the other variables? (Apologies, I tried to Google but just couldn't get the right search terms.) T = ( 1/AP - 1/P ) * Q I got this far, but don't know the next step(s): T/Q = 1/AP - 1/P Cheers. AI: $$T = \left(\frac{1}{AP} - \frac{1}{P}\right)\,Q=\frac{Q}{P}\left(\frac{1}{A}-1\right)\,.$$ Then $$P=\frac{Q}{T}\left(\frac{1}{A}-1\right)\,.$$
H: If $0 Sorry! This may appear an easy question for you but this is definitely not easy for me. Please suggest how to approach this problem. Thanks in advance. AI: Let $f(x,y,z)= \frac{\sin x+\sin y+\sin z}{\cos x+\cos y+\cos z}$ and its range is $$0=\lim_{z\to y\to x\to 0}f(x,y,z)< \frac{\sin x+\sin y+\sin z}{\cos x+\cos y+\cos z} < \lim_{x\to y\to z\to \frac\pi2}f(x,y,z)=\infty $$
H: A clarification on the definitions of Topological space I have a question regarding the following statement from the definitions of Topological Space. “ The intersection of a finite number of sets in T is also in T.” Here, what is the role of “finite” here? I do understand the proof of the statement but have no idea how the intersection of infinite sets behave. Edit: The statement seems to suggest that the intersection of infinite number of sets in T might not be in T. Is that even possible? Also, the statement is from https://mathworld.wolfram.com/TopologicalSpace.html Thanks AI: The sets in $T$ are like the "open" sets. So the axiom says that the intersection of finitely many open sets is open. You don't want to extend that to infinite intersections. For example in $\mathbb{R}$ each set $(-\frac{1}{n},\frac{1}{n})$ is open for all $n$. But the intersection of all of these is $\{0\}$, which is not open.
H: Use the notion of Integers to find a solution for $a+x=b$, $a,b,x \in \mathbb{N}$. In my syllabus for "Scientific Computing" it states the following idea. Consider the case $a \leq b$, $a,b \in \mathbb{N}$. This is essentially the same as $a + x = b$, $x \in \mathbb{N}$. Therefore, the only solutions to this equation are when $a$ is less than or equal to $b$. However, when we require for there to be a solution we need to introduce the idea of pairs, i.e. $(a,b)$ that describes $x$. However, it is clear to see that $(a,b)$ is not the only way to represent $x$. We also have $(a+1,b+1)$ which describes the same number. Therefore, we introduce equivalence relations that comes directly from the equation $a+x=b$: $(a,b) \sim (c,d) \iff a+d=b+c$. I'm just confused as to what this even means. How does $(a,b)$ describe $x$? If we have $1 \leq 5$, then the equation becomes $1+x=5 \Rightarrow x=4$. Then our equivalence relation becomes $(1,5)$ - so essentially the relation describes $x=b-a$? But then how does $(c,d)$ come into play? AI: From the Peano axioms you have constructed ${\mathbb N}=\{0,1,2,3,\ldots\}$ with $+$, $\times$, and $<\,$; and you have proven the usual rules of computing in this system. Now you want to enlarge ${\mathbb N}$ to the system ${\mathbb Z}$, which also contains the negative numbers, whereby the known rules are still valid. A simple idea would be to create for each $n\in{\mathbb N}_{>0}$ its negative clone $-n$, and put these clones into the bag. But it has turned out that proving all the rules now is a hopeless undertaking, because of all the "cases". Instead one uses a certain algebraic trick, and your quoted text tries to indulge in this trick. The idea is to represent all integer numbers, positive, $0$, and negative ones, by pairs of natural numbers. A pair $(a,b)\in{\mathbb N}^2$ then represents the solution $x$ to the equation $a+x=b$, i.e., the number that we usually call $b-a$. When $a\leq b$ then $b-a\in{\mathbb N}$ is already present in our world, and we would not need this pair-complication. But when $a> b$ then the pair $(a,b)$ is a new kind of thing, representing $a-b>0$ "unit steps backwards". Introducing these pairs creates a new difficulty: The same "new number" $x$, being $\geq0$ or $<0$, is represented by infinitely many pairs, e.g., $3=(2,5)=(16,19)$. This means that an actual "new number" is an equivalence class of such pairs. Some work is then needed to install good operations $+$ and $\times$ on the set ${\mathbb Z}$ of these classes, whereby these operations have to give the same result as the already present operations on the "old numbers" $\geq0$.
H: Is this function $L^1(]0,1])$? Background: I was asking myself if in order to prove that given two random variables $X,Y \in L^1$ (i.e. with finite mean value) then $\mathbb{E}[XY]=\mathbb{E}[X]\mathbb{E}[Y]$. I was asking myself if we can avoid to require $\mathbb{E}[\|XY\|]<+\infty$. I tried in this way: Define $\mathscr{Q}=]0,1] \times ]0,1] \subset \mathbb{R}^2$. Define $f(x,y)=\frac{1}{\sqrt{x}}\frac{1}{\sqrt{y}}$ for $(x,y) \in \mathscr{Q}$. My problem: How can I compute $\int_{\mathscr{Q}}f(x,y) d\mu$ where $\mu$ denote the Lebesgue measure of $\mathscr{Q}$? I think that $\int_{\mathscr{Q}}f(x,y) d\mu=+\infty$ but I cannot understand if it is true and I do not manage to proof correctly why. AI: For your particular example, $f\in L_1$. Use Fubini's theorem $$ \int_\mathscr{Q} f=\int^1_0\Big(\int^1_0\frac{1}{\sqrt{xy}}\,dx\Big)\,dy=\Big(\int^1_0\frac{1}{\sqrt{y}}\,dy\Big)\Big(\int^1_0\frac{1}{\sqrt{x}}\,dx\Big) $$ As for the added question, the answer is no in general but for sure if your random variables $X$ and $Y$ are independent. Example, take the probability space $(\mathscr{Q},\mathscr{B}),\mu)$ that you defined ($\mathscr{Q}=[0,1]^2$), $\mu$ Lebesgue measure on the plane. Define $X(x,y)=\frac{1}{\sqrt{x}}=Y(x,y)$ Then $X$ and $Y$ are integrable, by clearly $XY$ is not integrable and so $E[XY]\neq E[X]E[Y]$.
H: show that $\|x\|^{-(n-2)}$ is harmonic function let $f:\mathbb{R}^n \to \mathbb{R}$, and $f: \mathbf{x} \mapsto \Vert\mathbf{x}\Vert^{-(n-2)}$ show that $f$ is harmonic. I tried to take derivative by $x_1$ twice, and I got this result: $1/4 (-1 + 4 n^2) (x + x_{2...n})^{(-3/2 - n)}$ this of course doesn't sum up to zero. where am I wrong? thanks AI: Given $f(\mathbf{x})=\Vert\mathbf{x}\Vert^{-(n-2)}$ we need to show that $$\nabla^2 f =\sum ^{n}_{i=1}\frac{\partial^2 f}{\partial x_{i}^{2}} =0$$ Let's work out the first derivative: $$\frac{\partial }{\partial x_{i}} \\| \mathbf{x} \\| ^{-n+2} =( 2-n) \\| \mathbf{x} \\| ^{-n+1}\frac{\partial \\| \mathbf{x} \\| }{\partial x_{i}}$$ Recalling that $\\| \mathbf{x} \\| =\sqrt{\sum\nolimits ^{n}_{i=1} x_{i}^{2}}$, we can see that $$\frac{\partial }{\partial x_{i}} \\| \mathbf{x} \\| ^{-n+2}=( 2-n) \\| \mathbf{x} \\| ^{-n} x_{i}$$ Now, $$\frac{\partial^2 f}{\partial x_{i}^{2}}=\frac{\partial}{\partial x_i}\left( ( 2-n) \\| \mathbf{x} \\| ^{-n} x_{i}\right)$$ Using the product rule, $$\frac{\partial}{\partial x_i}\left( ( 2-n) \\| \mathbf{x} \\| ^{-n} x_{i}\right)=( 2-n)\left( -n\\| \mathbf{x} \\| ^{-n-2} x_{i}^{2} +\\| \mathbf{x} \\| ^{-n}\right)$$ $$=( 2-n) \\| \mathbf{x} \\| ^{-n-2}\left( -nx_{i}^{2} +\\| \mathbf{x} \\| ^{2}\right)$$ Thus $$\nabla^2 f = ( 2-n) \\| \mathbf{x} \\| ^{-n-2}\sum ^{n}_{i=1}\left( -nx_{i}^{2} +\\| \mathbf{x} \\| ^{2}\right)$$ $$=( 2-n) \\| \mathbf{x} \\| ^{-n-2}(-n\Vert\mathbf{x}\Vert^2+n\Vert\mathbf{x}\Vert^2)=0.$$
H: Real positive solution to specific system of equations In the course of an optimization problem, I have encountered the following system of equations for $\alpha, \beta > 0$: $$ \begin{align} a_1 + a_2 + b_1 + b_2 &= \alpha + \beta\\ a_1b_1 + \frac{1}{4} a_2 b_2 &= \alpha \beta. \end{align} $$ Is there a general approach to tackle such a system and does this system have a real solution with $a_1, a_2 , b_1, b_2 > 0$? AI: Note that $x + y = s$, $x y = t$ has solutions $$ x = \frac{s - \sqrt{s^2 - 4 t}}{2},\ y = \frac{s + \sqrt{s^2 - 4 t}}{2} $$ which are positive if and only if $s^2 \ge 4 t$, and $s,t>0$. Of course $(\alpha +\beta)^2 \ge 4 \alpha \beta$, with equality iff $\alpha = \beta$. If $\alpha \ne \beta$, we can choose any $a_2$ and $b_2$ that are positive but small enough that $s > 0$, $t> 0$ and $s^2 > 4 t$ where $s = \alpha + \beta - a_2 - b_2$ and $t = \alpha \beta - a_2 b_2/4$. Then we have positive solutions $a_1 = (s - \sqrt{s^2 - 4 t})/2$, $b_1 = (s + \sqrt{s^2 - 4 t})/2$. If $\alpha = \beta$, the system is not solvable with positive $a_1,a_2,b_1, b_2$, because squaring the first equation and subtracting $4$ times the second gives $$ \left( a_{{1}}-b_{{1}} \right) ^{2}+2\,a_{{1}}a_{{2}}+2\,a_{{1}}b_{{2 }}+{a_{{2}}}^{2}+2\,b_{{1}}a_{{2}}+a_{{2}}b_{{2}}+2\,b_{{1}}b_{{2}}+{b _{{2}}}^{2} =0$$
H: Calculate the following integral $\iint_{T}\frac{x^2\sin(xy)}{y}\,dx\,dy$ Calculate the following integral $$\iint_{T}\frac{x^2\sin(xy)}{y}\,dx\,dy\,,$$ where $$T=\{(x,y)\in\mathbb{R}^2:x^2<y<2x^2,y^2<x<2y^2\}$$ I found the $1/2\leq x\leq 1,1/2\leq y\leq 1$ but i got stuck on how set the limits of the integral AI: Your region is bounded by $y=x^2$, $y=2x^2$, $y=\sqrt x$, and $y=\sqrt{x/2}$. See the figure below. So you need to calculate the intersections. Intersection $y=x^2$ with $y=\sqrt x$: $$x^2=\sqrt x$$ This yields $x=0$ or $x=1$. At $x=1$ you have $y=1$ Intersection $y=x^2$ with $y=\sqrt{x/2}$ $$x^2=\sqrt\frac x2$$ This yields $x=0$ and $x=2^{-\frac13}\approx 0.79$. At this $x$ you have $y=2^{-\frac 23}\approx 0.63$ Intersection of $y=2x^2$ with $y=\sqrt{x/2}$ yields $x=0$ and $x=0.5$, with $y=0.5$ intersection of $y=2x^2$ with $y=\sqrt x$ yields $x=4^{-\frac13}=2^{-\frac23}$ and $y=2\cdot 4^{-\frac23}=2^{-\frac 13}$ This means that you can split the $x$ integral in three regions, from $1/2$ to $2^{-\frac23}$ to $2^{-\frac 13}$ to $1$. The corresponding limits for $y$ will be given by the curves that bound your region.
H: What is the Meaning of $Q[u_1] \cong Q[x]/\langle f(x)\rangle$? Suppose that $f(x) = ax^2 + bx + c \in Q[x]$ is irreducible. Then $\sqrt{(b^2 − 4ac)} \neq 0$, so that $f$ has two distinct roots - $$\begin{array}{*{20}c} {u_1 = \frac{{ - b + \sqrt {b^2 - 4ac} }}{{2a}}}{, u_2 = \frac{{ - b - \sqrt {b^2 - 4ac} }}{{2a}}} \\ \end{array}$$ I found in an article (similar examples can be found in other texts) - $$Q[u_1] \cong Q[x]/\langle f(x)\rangle.$$ Now I understand - $\langle f(x)\rangle$ is an ideal, $Q[x]/\langle f(x)\rangle$ is a quotient ring, elements of $Q[x]/\langle f(x)\rangle$ are cosets, i.e. $Q[x]/\langle f(x)\rangle$ is the set of polynomials in $Q[x]$ of degree less than $\text{deg}(f(x))$, i.e. degree $2$. On the other hand, $Q[u_1]$ is created by adjoining $u_1$, which is an element of degree 2 (the closed expression of $u_1$ has square root and square), to $Q_1$. Then how is it possible that $Q[u_1] \cong Q[x]/\langle f(x)\rangle?$ What does $Q[u_1] \cong Q[x]/\langle f(x)\rangle$ mean? What are the elements of $ Q[x]/\langle f(x)\rangle$ here? Plz explain elaborately with demonstration. Thanks. AI: The substitution homomorphism $\phi(X)\mapsto \phi(u_1)$ from $\mathbb{Q}[X]$ to $\mathbb{C}$ has kernel $\langle f(X)\rangle$ and image $\mathbb{Q}[u_1]$; so the result follows from the Isomorphism Theorem.
H: Why is $(1 - \frac{1}{n^{1-\epsilon}})^{n} < e^{-n^{\epsilon}}$ for $0 < \epsilon < 1$? This argument appears in one proof in my lecture and I don't know why this holds. Maybe someone knows a theorem that implies this inequality? Thanks for help. AI: If we take $x=e^{-n^\varepsilon}$ then the inequality to be proved is $$\left(1-\frac xn\right)^n<e^{-x}\tag1$$ where $x=n^{-\varepsilon}$. Taking logarithms, (1) is equivalent to $$\ln\left(1-\frac xn\right)<-\frac xn\tag2$$ at least as long as $n>x$. This is the same as $$\ln(1-y)<-y\tag3$$ for $y=x/n$. But (3) is valid for $0<y<1$; draw the graph and its tangent at $y=0$.
H: Find a homomorphism $f : H \to S_n$ such that $N = \ker(f)$. I am studying for an exam, and I got stuck on question on finding a certain homomorphism. For a group $G$, and a finite subset $A \subset G$, with $n=\#A$, consider subgroup $H=\{ g \in G \mid \text{for all }a \in A, gag^{−1}\in A\}$, and $N=\{g\in G\mid \text{for all }a\in A, gag^{-1}=a\}$ ($N$ is a normal subgroup of $H$). Then I have to find a homomorphism $f : H \to S_n$ such that $N = \ker(f)$. I do not really know how to start, in general I find it a bit hard grasp the concept of a function from a group to a permutation. Anyone that can help me get started on this? AI: The main clue here is that the subgroup $N$, which is the kernel of the map, is about conjugation, which is a standard action you will have learned about in your course. Actions are very relevant here, as an action of a group $H$ on a set $X$ is really just a map $H\rightarrow \mathrm{Sym}(X)$, and vice-versa (such homomorphisms define actions). Indeed, the map you are looking for correponds to the conjugation action, so we have the map $g\mapsto (a\mapsto g^{-1}ag, a\in A)$. Here, $(a\mapsto g^{-1}ag)$ defines a permutation of the set $A$, and so an element of $S_n$. The kernel is your subgroup $N$ as this permutation is the identity map permutation if and only if $g\in N$.
H: Formula about Closure of Sets I am reading the proof in link here in which I found a formula confusing. It says that ${\rm Cl}_X(W_0\cap A)={\rm Cl}_X(W_0\cap{\rm Cl}_X(A))$ due to $W_0$ is open. One direction ${\rm Cl}_X(W_0\cap A)\subseteq{\rm Cl}_X(W_0\cap{\rm Cl}_X(A))$ is trivial, but another direction seems not clear to me. In particular, the author mentions that this holds because $W_0$ is open. What is the role of $W_0$ here? I am confused about this formula. Hope anyone have good ideas on this. Solution. Inspired by the following answer, I have come up with a proof of that: For each $x\in{\rm Cl}_X(W_0\cap{\rm Cl}_X(A))$, let $U$ be an arbitrary neighborhood of $x$. Then $$U\cap(W_0\cap{\rm Cl}_X(A))=(U\cap W_0)\cap{\rm Cl}_X(A)\ne\emptyset.$$ As we can see, for each $y\in U\cap W_0\cap{\rm Cl}_X(A)\subseteq{\rm Cl}_X(A)$, since $U\cap W_0$ is a neighborhood of $y$ (That is where the openness of $W_0$ is used!), it follows that $$(U\cap W_0)\cap A=U\cap(W_0\cap A)\neq\emptyset.$$ Thus, we have $x\in{\rm Cl}_X(W_0\cap A)$, completing the proof. AI: Hint Suppose that $X=\mathbb R$ equipped with standard metric $\|\cdot \|$. Let $y\in Cl_X(W_0\cap Cl_X(A))$. Then, there is a sequence $(x_n)$ of $W_0\cap Cl_X(A)$ that converges to $y$. Since $(x_n)$ is in $Cl_X(A)$, there is a sequence $(x_{m}^n)_m$ in $A$ that converges to $x_n$. Since $x_m^n\to x^n$ when $m\to \infty $ and $W_0\ni x_n$ is open, there is $m_n\geq n$ s.t. $$\|x_{m_n}^n-x_n\|<\frac{1}{n}\quad \text{and}\quad x_{m_n}^n\in W_0,$$ for all $n$. Finally, $(x_{m_n}^n)$ is a sequence of $W_0\cap A$ that converges to $y$. Therefore $y\in Cl_X(W_0\cap A)$ as wished. Edit This proof can be very easily adapted to more abstract spaces. But since the question was rather on the necessity of $W_0$ to be open than on a topological proof of this statement, I put this answer as a Hint.
H: Given a finite set $X$, which of $\mathcal{P}(X\times X)\times\mathcal{P}(X\times X)$ and $\mathcal{P}({\mathcal{P}(X)})$ has more elements? This is a problem that I found in the book "Proofs and Fundamentals", by E. Bloch. Problem: Let $X$ be a finite set. Which of the two sets $\mathcal{P}(X \times X) \times \mathcal{P}(X \times X)$ and $\mathcal{P}({\mathcal{P}(X)})$ has more elements? Until now: I already proved that for a finite set $A$ with $a$ elements, $A \times A$ has $a^2$ elements and $\mathcal{P}(A)$ has $2^a$ elements. So, if $X$ has $n$ elements, then by the same argument, $\mathcal{P}(X \times X) \times \mathcal{P}(X \times X)$ has $2^{2n^2}$ and $\mathcal{P}({\mathcal{P}(X)})$ has $2^{2^n}$. I'm having trouble with showing which one is greater than the other, I tried induction but I didn't know how to finish. I also tried to see, for what values of $n$, one expression is bigger than the other, but I don't know how to solve that equation. The Point: I would really appreciate if someone could explain how can I deduce which one is greater than the other, and, if possible, to explain other methods that are suitable to solve this problem. Thank you very much for your attention AI: Its true if $n=7$. So lets try induction with this base case. From the induction hypothesis we get: $2^{n+1}=2^n+2^n\geq 2n^2+2n^2$. OTOH: $2(n+1)^2=2n^2+4n+2$. So we need: $2n^2\geq 4n+2$. Again this is true if $n=7$. So try this one by second induction. Here the induction step is easier. If $2n^2\geq 4n+2$ then: $2(n+1)^2=2n^2+4n+2\geq 4n+2+4n+2\geq 4(n+1)+2$.
H: Why not to calculate the limit of individual terms when the number of terms approach infinity due to limiting variable? My instructor was using this example to illustrate his point- $$ \lim_{n\to \infty} \frac{1+2+3+...+n}{n^2}$$ The numerator is an AP and the limit can be easily calculated this way to be a finite number($0.5$ in this case). But, if we use the sum law and separate all the $\frac{r}{n^2}$ terms, the individual limits will be $0$ and so will be the answer. This is what my instructor told me to avoid. My question is- Why can we not do this? Should we also not do this if the individual limits are finite values? Is this specific to the sum law? AI: The sum law, or any of the other limit laws, is usually stated only for the case of 2 terms. By induction, we may extend it to 3, 4, 5, or any other fixed finite number of terms, where fixed means that the number does not depend on the limit variable $n$. However, as your example shows, we may not extend it to the case where the number of terms depends on $n$.
H: Prove that the integral $\int_a^b \frac{\sin(x)}{x}dx$ is bounded uniformly. How do I show that exists a constant $M>0$ such that, for all $0\leq a \leq b < \infty$, $$\left\|\int_a^b \frac{\sin(x)}{x}dx\right\| \leq M.$$ I just read on Richard Bass book's that is enough to prove the uniformly boundeness of the integral $\int_0^b \frac{\sin(x)}{x}dx$, but I can't see why... AI: Let $G(x)=\int^x_0\frac{\sin t}{t}\,dt$. If One proves that $\|G(x)\|\leq M$ for some $M>0$ and all $x\geq0$, then $$\|G(b)-G(a)\|\leq \|G(b)\|+\|G(a)\|\leq 2M$$ Hint: $\lim_{x\rightarrow+\infty}G(x)$ exists (it is $\frac{\pi}{2}$ but the value for this exercise is not that important) If you know this, then the problem is almost done. If you don't, then you may try to write $G(x)$ as sum of decreasing alternating terms just to show that indeed $\lim_{x\rightarrow\infty}G(x)$ exists. For instance for $n\pi\leq x<(n+1)\pi$, $$G(x)=\int^x_{n\pi}\frac{\sin t}{t}\,dt +\sum^n_{k=1}\int^{n\pi}_{(n-1)\pi}\frac{\sin t}{t}\,dt $$ The first term is bounded by $\pi$, the second is a sum of terms $\sum^n_{k=1}a_k$ which alternate in sign but that in absolute value $\|a_n\|\xrightarrow{n\rightarrow\infty}0$.
H: Integrality and normality of ideals Consider a ring $R$, an idela $I$. An element$z\in R$ is integral over $I$ if $z$ satisfies the equation $$z^n+a_1z^{n-1}+\ldots+a_{n-1}z+a_n=0,$$ where $a_i\in I^i$. We define the integral closure $\overline{I}$ of an ideal as the set of elements in $R$ which are integral over $I$ (and this is an ideal of $R$). We say that $I$ is integrally closed if $I=\overline{I}$, and that $I$ is normal if all the powers $I^k$ are integrally closed (I'm using Villarreal's Monomial Algebras). Question: I'm a bit confused, because from my point of view $I$ is integrally closed if and only if $I$ normal. Proof: If $I$ is normal, then all the powers $I^k$ are integrally closed, hence imposing $k=1$ implies $I$ is integrally closed (maybe here I'm cheating a bit, but Villareal doesn't say anything like $k>1$. On the other hand, if $I$ is integrally closed, then $I=\overline{I}$ by definition, hence $I^k=\overline{I}^k$ for any $k$. I strongly suspect there's something wrong (strange to give $2$ different names for the same object), but I can't spot the mistake and I'd like to understand! AI: You say "On the other hand, if I is integrally closed, then $I=\bar{I}$ by definition, hence $I^k=\bar{I}^k$ for any k.", but $\bar{I^k} \ne \bar{I}^k$ in general so one cannot conclude from this that $I$ is normal. In fact, integrally closed ideals are frequently not normal. For a simple example, take $R=k[\![x,y]\!]/(x^2,xy)$ and set $\mathfrak{m}=(x,y)$. Then of course $\mathfrak{m}$ is integrally closed, but, when $k>1$, $\mathfrak{m}^k=(y^k)$. As the integral closure of any ideal always contains the nilradical of $R$, $x$ must be in the integral closure of any ideal, and so, in particular, $\mathfrak{m}^k$ is not integrally closed when $k>1$.
H: Standard etale maps are etale We define a finitely presented $R$-algebra $A$ to be etale if for every $R$-algebra $S$, every ideal $I\subset S$ such that $I^2=0$ and every homomorphism $$\phi: A\to S/I$$ there exists a unique homomorphism $$\psi: A\to S$$ such that $\pi\circ\psi=\phi$, where $$\pi: S\to S/I$$ is the standard projection map. We call an $R$-algebra standard etale if it is isomorphic to $R[X]_g/(f)$ for some $f,g$ such that $f'$ is invertible in $R[X]_g/(f)$. It is not obvious to me that a standard etale algebra is etale. How would one show this, or where would one start? AI: Let $S$ be any $R$-algebra, and $A=R[X]_g/(f)$ be any standard étale algebra. The $R$-morphisms $A \rightarrow S$ are given by the $x \in S$ such that $f(x)=0$ and $g(x) \in S^{\times}$. Now let $S$ be any $R$-algebra, and $I \subset S$ an ideal with square zero. We need to show that for each $x \in S$ such that $f(x) \in I$ and $g(x)$ is invertible mod $I$ ($x\,\mathrm{mod}\,I$ represents one morphism $A \rightarrow S/I$), there is a unique $x' \in S$ congruent to $x$ mod $I$ such that $g(x') \in S^{\times}$ and $f(x')=0$. Note that since $I^2=0$, $y \in S$ is invertible iff it is invertible mod $I$. So we need to show that there is a unique $x' \in x +I$ such that $f(x')=0$. But since $I^2=0$, it is easy to see that for $y \in I$, $f(x+y)=f(x)+f'(x)y$. To conclude, one only needs to note that $f'(x) \in S^{\times}$, which is easy since $f' \in A^{\times}$ as $A$ is standard étale.
H: Existence of Euler path in $K_5$, the complete graph with five vertices Construct an Eulerian path in $K_5$ I tried with the aid of the theorem: A graph contains an Eulerian path if and only if there are at most two vertices of odd degree. But I became stuck while ending the walk at initial. AI: If you put the five points as the vertices of a regular pentagon, the graph is a pentagon plus a five pointed star. Start at any vertex, draw the pentagon to get back to where you started. Now draw the five pointed star. When you get back the second time, you have your path. If the vertices are $ABCDE$ clockwise, you have $ABCDEACEBDA$
H: How Lebesgue integration solved the problem of a function being integrable but its limit is not integrable? My professor gave us the following form of Dirichlet function as an example of the problems we faced in Riemann integration: $\{r_{n}\}$ enumeration $\mathbb{Q} \cap [0,1]$ $$ f_{n}(x) = \begin{cases} 1 & \quad x \in \{r_{1}, ... , r_{n}\} \\ 0 & \quad \text{otherwise}. \end{cases} $$ And he said that: each $f_{n}$ is integrable but its limit is not integrable. My questions are: 1- why each $f_{n}$ is integrable but its limit is not integrable? 2- How did Lebesgue integration solve this problem? Could anyone help me understand answers to these questions, please? AI: Each $f_n$ is integrable because you can check that each lower sum is equal to $0$ and that the upper sums can take (positive) values as small as you want. Therefore, the (Riemann) integral of $f$ is $0$. However, if $f$ is the limit, then every lower sum is $0$ and every upper sum is $1$. Therefore, $f$ is not Riemann-integrable. In the case of Lebesgue integration, $f$ is integrable, and its integral is $0$.
H: Regarding closed half spaces in $\mathbb{C}^n$ I know that a closed convex set $\Omega\subset \mathbb{R}^n$ is the intersection of the half spaces containing it. Here a half space is of the form $H=\{x\in: \mathbb{R}^n: f(x)\geq0 \}$, where $f: \mathbb{R}^n \longrightarrow \mathbb{R}$ is an affine map. And $f$ is an affine map if there exists a linear map $h: \mathbb{R}^n \longrightarrow \mathbb{R}$ and a fixed vector $c\in \mathbb{R}$, such that $f(x)=h(x)+c$ for every $x\in \mathbb{R}$. Now I am reading a paper where a closed half space in $\mathbb{C}^n$ is denoted by $H’=\{z\in: \mathbb{C}^n: a(z)\geq0 \}$ for some affine function $$a:z=(z_k)\mapsto b+ \sum \alpha_kz_k+ \sum \overline{\alpha_kz_k},$$ where $b\in\mathbb{R}, \alpha_k\in\mathbb{C}$. Can anyone tell why here $b\in \mathbb{R}$? And why is the expression of $a$ defined as such, I do not see it as a sum of a linear function and a complex scalar. Can anyone suggest a reference where closed convex sets in $\mathbb{C}^n$ are defined and explained this way? AI: The point is that you need your function $a$ to be real-valued in order for $a(z) \ge 0$ to make sense. So you take the real part of the sum of a linear function and a scalar. $$ \text{Re}(\beta + \sum_{k} \alpha_k z_k) = \text{Re}(\beta) + \frac{1}{2} \sum_k \alpha_k z_k + \frac{1}{2} \sum_k \overline{\alpha_k z_k}$$
H: A doubt regarding the delta complex structure for a circle I'm currently self-learning algebraic topology from Hatcher. In page 106, Hatcher computes $H_{n}^{\Delta}(S^1)$. To do this, he says "let $X = S^1$ with one vertex and one edge". The diagram given in the book is clear but I'm not able to see how this is a $\Delta$-complex for a circle. Don't you need 2 1-simplexes, with their endpoints indentified to get a $\Delta$-complex for a circle? In this case you end up with 2 edges and 2 vertices. I apologize in advance if this is a trivial question. AI: The definition that Hatcher uses for $\Delta$-complexes is sufficiently flexible to allow for this. The point is that if $\sigma_\alpha$ is a map of an $n$-simplex with a face $K$, the definition demands that $\sigma\|_{K}$ is the same as a characteristic map $\sigma_\beta$ for some $n-1$-simplex. In particular, the maps for different faces need not be distinct (this is a difference between $\Delta$ and simplicial complexes). So we are allowed to have both edges of a $1$-simplex be mapped to the same point (and get a representation of $S^1$ as $[0,1]/(0 \sim 1)$).
H: Harmonic series "fulfills" Cauchy criterion Let $S_n=\sum_{k=1}^n \frac{1}{k}$ the $n$-th partial sum in the harmonic series. Once can prove that $$ \lim_{n\rightarrow\infty}\|S_{n+p}-S_n\|=0, $$ for all $p\in\mathbb{N}$. On the other hand, the harmonic series diverges. So the question is why the above limit "fulfills" the Cauchy criterion for series, but the harmonic series still diverges? The Cauchy Criterion states that for a sequence $(a_n)$ in the complex numbers, $\sum_n a_n$ converges if and only if to each $\epsilon>0$ there is a $n_0\in\mathbb{N}$ such that $q>p\geq n_0$ implies $\|\sum_{n=p+1}^q a_n\|=\|S_q-S_p\|<\epsilon$. AI: You know that for all $p$ and $\varepsilon$ there is some $N_{p,\varepsilon}$ such that, for all $n\ge N_{p,\varepsilon}$, $\lvert a_{n+p}-a_n\rvert\le \varepsilon$. What you do not know is wheter or not for all $\varepsilon$ there is some $N_\varepsilon$ such that, for all $n\ge N_\varepsilon$ and for all $p$, $\lvert a_{n+p}-a_n\rvert\le\varepsilon$. As of why these need not be equivalent, the instance at hand shows.
H: Why Folland's Advanced Calculus is so strict about uniform convergence? Folland's Advanced Calculus uses uniform convergence to justify the interchange of limits (i.e. to change order of integration and summation). But actually uniform convergence is far powerful than only justifying such an action. For example Fubini's theorem states a sufficient condition to interchange the order of limits and this condition is far weaker than uniform convergence. Is it pedagocical thing or mentioned stronger theorems need much background than the level of Advanced Calculus? Even integrating (Riemann) the geometric series term by term $$\sum_{n=0}^{\infty}x^n$$ cannot be justified via uniform convergence since it is not uniformly convergent on $(0,1)$ since the series grows unboundedly near $1$. But Fubini's theorem states that if $f_n(x) \geq 0$ for all $x\in(0,1)$ and for all $n\in\mathbb{N}$ then the series can be integrated term by term. Why there are questions like this in Folland. I get a bit frustrated when I cannot justify the actions I conduct, properly. AI: Fubini's theorem is a deep result of measure theory. Folland probably uses uniform convergence in order to stay in sync with the background of the reader. The geometric series does not converge uniformly, but it converges locally uniformly (or uniformly over compact subsets, if you prefer).
H: Why does $\sum_{i=n}^{2n-1}\binom{i-1}{n-1}2^{1-i}$ computes the probability of $n$ head or tails $\sum_{i=n}^{2n-1}\binom{i-1}{n-1}2^{1-i}$ For $i = n,n+1,\ldots, 2n - 1$, the sum above computes $P(E_i)$, the probability that i tosses of a fair coin are required before obtaining $n$ heads or $n$ tails. I am not asking for a solution. I am confused by how that summation computes the probability described in the hint. Shouldn't the binomial coefficient be $\binom{i}{n}$? That's the only question I have. AI: The earliest that you can get $n$ heads or $n$ tails is the $n$-th toss, and the latest is the $(2n-1)$-st toss. How many ways are there to get it on the $i$-th toss, where $n\le i\le 2n-1$? Let’s say that you get the $n$-th tail on the $i$-th toss. That means that you must have got $n-1$ tails in the first $i-1$ tosses, and they could any $n-1$ of those tosses. Thus, there are $\binom{i-1}{n-1}$ strings of heads and tails that would allow you to get the $n$-th tail on the $i$-th toss. The probability of getting any one of those strings is $2^{-(i-1)}=2^{1-i}$. The probability of getting a tail on the $i$-th toss is $\frac12$, so the probability of a string of length $i$ that has exactly $n$ tails and ends with a tail is $\frac12\cdot\binom{i-1}{n-1}2^{1-i}$. The probability of getting the $n$-th head on the $i$-th toss is identical, so the probability of getting one or the other of these outcomes is $$2\cdot\frac12\binom{i-1}{n-1}2^{1-i}=\binom{i-1}{n-1}2^{1-i}\;.$$
H: Proving the open unit disc in $\Bbb R^2$ is open I'm going through Rudin chapter two on Topology, and have been working on open sets, closed sets and material related to that. I understand the concepts, but am struggling to apply it to some examples. In the book it has the following: The set of all complex numbers $z$ such that $\|z\| < 1$ is Open and Bounded, but not Closed and not Perfect. I'm trying to prove each to get practice with the definitions. But am stuck (so far I've only tried proving it is open) Open: Let $D = \{(x,y) \in \mathbb{R}^{2} : \sqrt{x^{2} + y^{2}} < 1\}.$ I need to show that every point in $D$ is an interior point i.e. if I draw a ball around a point in $D$ that ball lies completely inside $D$. My idea was to take $\epsilon = 1 - \sqrt{(s)^{2} + (t)^{2}}$ i.e. the segment that connects the point $(s,t)$ to the boundary of $D$. Let $x = (x_{1}, y_{1}) \in D$ and let $a = (a, b) \in D.$ Then $B_{\epsilon}(a) = \{x \in D: d(x,a) < \epsilon\}.$ I get a bit stuck now. I need to apply the definition of the ball and probably the triangle inequality (I tried the 1D case a simplified example). $ \sqrt{x_{1}^{2} - a^{2}} + \sqrt{y_{1}^{2} - b^{2}} < \epsilon$ Any help appreciated. Thanks. AI: Take $(x,y)\in D$ and let $\varepsilon=1-\sqrt{x^2+y^2}$. If $(s,t)\in B_\varepsilon\bigl((x,y)\bigr)$, then\begin{align}d\bigl((s,t),(0,0)\bigr)&\leqslant d\bigl((s,t),(x,y)\bigr)+d\bigl((x,y),(0,0)\bigr)\\&<1-\sqrt{x^2+y^2}+\sqrt{x^2+y^2}\\&=1.\end{align}In other words, $(s,t)\in D$.
H: Compact and $T_2$ - space question Give $X$ is compact $T_2$-space and $(F_n)$ is the sequence of closed sets that $F_n \supset F_{n+1}$. I have proved that $\displaystyle F=\bigcap_{n=1}^{+\infty}F_n \neq \varnothing$. So, i feeling that if $G$ is open contain $F$ then exists $n_0 \in \mathbb{N}$ that $F_{n_0} \subset G$. I have tried to use contracdiction to prove it but i failed. I wondering that my feeling is right or not? AI: From the hint above of Prof. Brian M. Scott, I have proved this problem as follows. Suppose that $F_n \not\subset G, \forall n \in \mathbb{N}$. Fix $H_n=F_n \setminus G \neq \varnothing$. We claim that $H_n$ is closed and $H_n \supset H_{n+1}$ for all $n\in \mathbb{N}$. Since $H_n \supset H_{n+1}, \forall n \in \mathbb{N}$ then $\displaystyle \bigcap_{j \in J}H_j \neq \varnothing$ for all $J$ is finite in $\mathbb{N}$. Moreover, since $X$ is compact space then $\displaystyle \bigcap_{n=1}^{+\infty}H_n \neq \varnothing$ (). Other side, $\displaystyle \bigcap_{n=1}^{+\infty}H_n= \bigcap_{n=1}^{+\infty}(F_n\setminus G)= F\setminus G = \varnothing$ (contradict with ()). So, there exists $n_0$ that $F_{n_0} \subset G$.
H: $(\lvert a\rvert +\lvert b\rvert)^{p}\leq 2^{p}(\lvert a\rvert^{p} +\lvert b\rvert^{p})$ for $p > 1$ and are absolute values necessary? Is it true that, for any $a,b\in \mathbb R$ and $p \geq 1$, we have $(\lvert a\rvert +\lvert b\rvert)^{p}\leq 2^{p}(\lvert a\rvert^{p} +\lvert b\rvert^{p})$ If $p=2$ we have an explicit way to calculate by expansion, $(\lvert a\rvert +\lvert b\rvert)^{2}=\lvert a\rvert^2+2\lvert a\rvert \cdot \lvert b\rvert +\lvert b\rvert^2 \leq 4\lvert a\rvert ^{2}+4\lvert b\rvert^{2}=2^2(\lvert a\rvert ^{2}+\lvert b\rvert^{2})$ How can I go about proving the others? Is positivity necessary here? AI: $(\|a\|+\|b\|)^p\leq (2\max\{\|a\|,\|b\|\})^p=2^p(\max\{\|a\|,\|b\|\})^p\leq 2^p(\|a\|^p+\|b\|^p)$ The last step simply follows from the fact that $(\max\{\|a\|,\|b\|\})^p$ is either $\|a\|^p$ or $\|b\|^p$, so it is definitely not bigger than their sum. Here we indeed used that they are nonnegative.
H: How to convert coeffecient multiplying $t$ in a trig function (e.g. $\sin(bt)$) into Hertz? How to convert coeffecient multiplying $t$ in a trig function (e.g. $\sin(bt)$) into Hertz? For example, if I want a machine to produce waves in my bathtub such that the waves follow the function $\sin(4.224 t)$, but the machine operates in Hertz (Hz), how do I make this conversion? The period of $\sin(4.224 t)$ is $2\pi / 4.224$, and so the frequency is $4.224/ 2\pi$. I know that much, but I'm not sure how to convert this frequency (which seems to be unitless) to Hz... From searching online I've found an equation for "angular frequency", $w=2\pi f$, but I'm not sure what the $w$ and $f$ would be in this context (also not sure if this is even relevant/usable in this situation) AI: Hertz is $s^{-1}$ in other words "per second". Sensibly, the argument to $\sin$ will be unitless but assuming that $t$ is in seconds then $b$ should be in $s^{-1}$. For frequency, you want whole waves per second, so considering that the period of $\sin$ is $2 \pi$, the frequency will be $f = \frac{b}{2 \pi}$. This is also $s^{-1}$ but Hz is traditionally used. The angular speed $\omega$, which is just $b$ for you is larger, by a factor of $2 \pi$, because it is measured in radians per second rather than whole revolutions. This Wikipedia article uses angular frequency, angular speed,or a few other terms and reserves angular velocity for the vector. This is consistent with linear motion. The main distinction that I am trying to make is between measuring whole revolutions or radians. In day to day life, rotational rate is typically measured in revolutions per time unit. E.g. records have typical speeds of $33 \frac{1}{3}$, $45$, or $78$ rpm and tachometers in cars also use rpm (or a multiple of it). However, if you see $\sin(\omega t)$ then $\omega$ will be in radians per unit time; probably second but it could be other units as long as they match $t$. $\omega t$ needs to be dimensionless and in radians. I guess that rotational rate could be measured in degrees per unit time but I have never seen that. Radians are often treated as if they were a unit but they are not really. They are not arbitrary in the way that the metre or the yard is. A measurement in radians is just a number. So, radians per second is actually just $s^{-1}$ and we could use Hz. However, because other measures of angles are common, degrees or revolutions, we generally feel the need to emphasize that radians are being used. Hz is usually reserved for whole waves per second. Becquerel is also $s^{-1}$ but it is used for radioactive decays per second. More units are used than are really necessary.
H: In $\triangle ABC$, $D$ bisects $AB$, $E$ trisects $BC$ and $\angle ADC=\angle BAE$. Find $\angle BAC$ Giving a $\triangle ABC$ having point $D$ as the mid point of $AB$ and $E$ as the trisection point on $BC$ such that $BE>CE$. If $\angle ADC=\angle BAE$. Find $\angle BAC$ I can't solve this problem but I was able to construct the picture of it and it seems that $\angle BAC=90^\circ$. In the picture below is what I could find so far ( there is a tons of parallel but I don't add it to make to picture more comfortable). Any idea? AI: Let X be the intersection of AE and CD. Evaluate $$\frac{DX}{XC}= \frac{Area_{ADE}}{Area_{AEC}}=\frac{\frac12\frac23 Area_{ABC}}{\frac13 Area_{ABC}}=1 $$ Then, with $\angle ADC=\angle BAE$, we get $AX= DX=CX$. So, $A$, $D$ and $C$ are cyclic with $X$ being the circumcenter. Thus, $\angle BAC = 90^\circ$.
H: Orbit Space of Action of $SO(n)$ on $\mathbb{E}^n$ I'm trying to solve problem 28, chapter 4 of M. A. Armstrong's Basic Topology : Describe the orbits of the natural action of $SO(n)$ on $\mathbb{E}^n$ as a group of linear transformations, and identify the orbit space. I take the norm function as an identification map and try to use the fact that $\mathbb{E}^n$ with this identification is homeomorphic to $[0,1)$. I know each orbit of action $SO(n)$ on $\mathbb{E}^n$ is the same as an equivalence class of identification space but I need to show that the identification space and orbit space of acting $SO(n)$ on $\mathbb{E}^n$ are homeomorphic to claim the orbit space is homeomorphic to $[0,1)$. Can anyone help me to show that? AI: Perhaps if I unwind the notation of this problem for you, it will be more evident what you have to do. Let me introduce some notation for the norm function, say, $$N(\vec p) = \\|\vec p\\| = \sqrt{p_1^2 + \cdots + p_b^2} $$ What you have to prove is that the set of orbits of the action of $SO(n)$ on $\mathbb E^n$ is equal to the set of subsets of the form $$S_r = \{\vec p \in \mathbb E^n \mid N(\vec p) = r\}, \quad r \in [0,\infty) $$ Of course, if $r>0$ then $S_r$ is the sphere of radius $r$ centered on the origin, and $S_0 = \{\vec 0\}$. The next main step is to prove is that the orbits of the action of $SO(n)$ on $\mathbb E^n$ are precisely the spheres $S_r$ of radius $r>0$ and the set $S_0$, and I'll assume that you know how to do that. The final step involves applications of concepts of quotient maps and quotient topologies. One must check that: The norm function $N : \mathbb R^n \to [0,\infty)$ is a quotient map. Once that has been checked, one must then apply The universal property of quotient maps. It follows that $N$ induces a homeomorphism from the quotient topology on the orbit space of the action to the space $[0,\infty)$. If these things are unfamiliar to you, you'll have to learn about quotient maps and quotient topologies. I suggest a good topology book, such as Munkres "Topology".
H: Simplifying $\ln{U}=-2\sin x+\ln{\cos x}+\ln{C}$ to find $U$. I need to simplify this expression and find my $U$. What should I do? $$\ln{U}=-2\sin x+\ln{\cos x}+\ln{C}$$ AI: It is $$U=e^{-\sin(x)+\ln(C\cos(x))}$$
H: Inequality proof including $n!$ $n^{n}e^{-n+1} \le n! \le n^{n}e^{-n+1} n$, $n \in \mathbb{N}$ I'm struggling solving the inequality above, I have tried AM-GM, Bernoulli but I guess now that the proof is based maybe on induction.The squeeze theorem can make maybe also sense. I appreciate any help. AI: $1. \log$ and $\exp$ are monotonoe increasing functions, so if $\log x < \log y \to x <y$, same for $\exp$. $2.n! = e^{\sum_{k=1}^{n}\log k}$ $3. \int_{1}^{n} \log x dx = n \log n -n +1 $ Can you do the rest?
H: Suppose $a$, $b$, $c$, and $d$ are real numbers, $00$. Prove that if $ac\geq bd$ then $c>d$. Not a duplicate of Prove that if $a$, $b$, $c$, and $d$ are real numbers and $0 < a < b$ and $d > 0$ and $ac ≥ bd$ then $c > d$ This is exercise $3.1.11$ from the book How to Prove it by Velleman $(2^{nd}$ edition$)$: Suppose $a$, $b$, $c$, and $d$ are real numbers, $0<a<b$, and $d>0$. Prove that if $ac\geq bd$ then $c>d$. The solution to this problem is available in appendix $1$, but I was wondering about another possible solution: Proof. We will prove the contrapositive. Suppose $c\leq d$. Since $0<a<b$, then multiplying $a$ by $c$ and $b$ by $d$ will preserve the direction of the inequality and so $ac<bd$ as required. Therefore if $ac\geq bd$ then $c>d$. $Q.E.D.$ Is the above proof valid? AI: I'm not happy with "then multiplying a by c and b by d will preserve the direction of the inequality". I'm not sure that's clear and I'm not sure you can claim that without justification. You could prove that as a lemma. Lemma: If $m < n$ and $0 < u \le v$ the $mu < nv$. Pf: $mu < nu$ and $nu \le nv$ (as well as $mu \le mv$ and $mv < nv$ ) so by transitivity $mu < nv$.
H: Quotient lattice: definition, references I apologize in advance for this dumb question, but I don't know where to look and I'd like to understand. Consider the lattice $\mathbb{Z}^n$ (for me a lattice is a free abelian group of finite rank), and consider an element $v=(v_1,\ldots,v_n)\in\mathbb{Z}^n$. Consider the quotient $\mathbb{Z}^n/\mathbb{Z}v$: I want to prove this is a lattice of rank $n-1$. Let us first understand what the quotient is: given $x,y\in\mathbb{Z}^n$, $x=(x_1,\ldots,x_n)$, $$x\sim y \iff x_1-y_1=tv_1,\hspace{0.3cm}\ldots,\hspace{0.3cm} x_n-y_n=tv_n,$$ where $t\in \mathbb{Z}$. At this point I get stucked: I can isolate $t=\frac{x_i-y_i}{v_1}$ and put it in the remaining $n-1$ equation, but I don't see how this is still a lattice. I tried to do a small example, like $\mathbb{Z}^2$, $v=(2,3)$, then $\mathbb{Z}^2/\mathbb{Z}v$ has the relation \begin{equation} \begin{cases} x_1-y_1=2t \\ x_2-y_2=3t, \\ \end{cases} \end{equation} hence $x\sim y \iff 3x_1-2x_2=3y_1-2y_2$. But how to continue from this? I guess $$\mathbb{Z}^2/\mathbb{Z}v\simeq \mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z},$$ but I don't see why this is a lattice, and of rank $1$. I know my question is rather vague, but I really struggle with this construction and I need to understand it. Any help would be much appreciate. AI: In your example, let $w=(1,2)$. Then $\mathbb{Z}^2 = \mathbb{Z}v \oplus \mathbb{Z}w$ and so $$\mathbb{Z}^2 / \mathbb{Z}v = (\mathbb{Z}v \oplus \mathbb{Z}w)/(\mathbb{Z}v \oplus 0) \cong \mathbb{Z}w \cong \mathbb{Z}$$ This exact argument only works in the general case when the entries on $v$ are coprime.
H: Does $\iint_{\mathbb{R}^2}\frac{\sin(x^2+y^2)}{x^2+y^2}\,dx\,dy$ converge? Consider $$\iint_{\mathbb{R}^2}\frac{\sin(x^2+y^2)}{x^2+y^2}\,dx\,dy$$ My try: changing to polar coordinates and then calculate the integral $$\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\sin(r^2)}{r^2}\,dr\,d\theta\underset{r^2=u}{=}\pi\int_{0}^{\infty}\frac{\sin(u)}{u}\,du,$$ which converges but I saw in the solutions of a test that this integral diverges. Was I wrong? AI: As you suggested, let's change it to polar coordinates: $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{\sin(x^2+y^2)}{x^2+y^2}\,dxdy$$ $$=\int_{0}^{2\pi}\int_{0}^{\infty}\frac{\sin(r^2)}{r^2}\,rdr d\theta$$ $$=\pi\int_{0}^{\infty}\frac{\sin(r^2)}{r^2}\,2rdr$$ Let $t=r^2$, $dt= 2r\,dr$: $$ \Rightarrow \pi\int_{0}^{\infty}\frac{\sin(t)}{t}\,dt $$ This integral is convergent and its value is $\pi/2$; see, for instance, https://dlmf.nist.gov/5.9.7 or Evaluating the integral $\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$?. So the final answer is ${\pi^2}/{2}$.
H: Convergence of simple non-negative functions to a measurable function The question is from "Measure theory and probability theory (pag. 51)" by Krishna and Soumendra that lacks of a demonstration. Let $(\Omega, \mathcal{F}, \mu)$ be a measure space and $f:\Omega \rightarrow [0, \infty]$ a non-negative measurable function. Suppose $\{\delta_n\}$ to be a sequence of positive real numbers and $\{N_n\}$ a non-decreasing sequence of integers such that $\delta_n \rightarrow 0$, $N_n \rightarrow \infty$ and $\delta_n N_n \rightarrow \infty$ (for example $\delta_n=2^{-n}$ and $N_n=n2^n$). Define \begin{gather} f_n(\omega)=\left\{ \begin{array}{ll} j\delta_n \ \ \text{if} \ \ j\delta_n \le f(\omega) <(j+1)\delta_n \ \ j=0,\dots,N_n-1 \\ \delta_nN_n \ \ \text{if} \ \ f(\omega) \ge \delta_nN_n \end{array} \right. \end{gather} I need to prove that $\{f_n\}$ converges to $f$ pointwise and that each $f_n$ is measurable. AI: The preimages of $f_n$ are unions of sets of the form $\{\omega : a \le f(\omega) < b\}$ and $\{\omega : f(\omega) \ge a\}$, which are measurable sets because $f$ is measurable. For a fixed $\omega$, we want to check that $f_n(\omega) \to f_n(\omega)$. Because $\delta_n N_n \to \infty$, we have $f(\omega) < \delta_n N_n$ for all large $n$. By the definition of $f_n$, we have $f_n(\omega) = j\delta_n$ for some $j$, so $f_n(\omega) \le f(\omega) < f_n(\omega) + \delta_n$. Taking $\delta_n \to 0$ establishes the convergence.
H: Can I deduce from $ \frac{a}{b} = \frac{x}{y} $ that a = x and b = y where $ a, x \in Z $ and $ b, y \in Z^{+} $? Can I deduce from $ \frac{a}{b} = \frac{x}{y} $ that a = x and b = y where $ a, x \in Z $ and $ b, y \in Z^{+} $? I'm wanting to prove that the function $ f(x, y): \frac{x}{y} $ where $(x , y) \in \mathbb{Z} \times \mathbb{Z} ^{+} $ is injective. I'm using the definition of injectivity: $ \forall a, b \in X, f(a) = f(b) \Rightarrow a = b $ but I can't seem to find a way to proof that $f(x, y) = f(x', y') \rightarrow (x = x' \wedge y = y') $ AI: Nope: $\frac ab = \frac {ak}{bk}$ but we can not conclude that $ak =a$ nor that $bk=b$. For example $\frac 34 = \frac 68$ but $3\ne 6$ and $4\ne 8$. But if $\frac ab = \frac xy$ and both are "in lowest terms" you can conclude that $a=x$ and $b=y$. And if $\frac ab$ is in lowest terms but we don't know if $\frac xy$ is (or if we know it is not) the we know thre is an $m\in\mathbb Z$ so that $am = x$ and $bm = y$. (And we know that $m = \frac xa = \frac yb$). And if neither are necessarily in lowest terms you can conclude that there is a value $k\in \mathbb Q$ so that $ak = x$ and $bk = y$. (we can figure that $k = \frac xa =\frac yb$. And we can figure that if $\frac ab$ is in lowest terms then $k$ is an integer.) .... A case where we might fall into trouble might be a problem: Suppose $x = n + 3$ and $y = n + 6$ and $\frac xy = .75$ what is $n$. If you figure $\frac xy = .75 = \frac 34$ so $x=3$ and $y=4$ so $n+3 =3$ and $n+6=4$ so $n=0$ and $n=-2$ then... well, you just stapled yourself six ways to sundays. But instead you $\frac xy = \frac 34$ so $x = 3m$ and $y = 4m$ for some integer $m$ ($\frac 34$ is in lowest terms so we can assume $m$ is an integer). So $n+3 = 3m$ and $n+6 = 4m$ so $n = 3m -3$ and $n = 4m -6$ so $3m-3 = 4m-6$ so $6-3 = 4m -3m$ or $3 = m$ so $m=3$ and $n+3 = 33=9$ so $n =6$ and $n+6=43 =12$ so $n = 6$. That's the correct answer. But, of course, the rule $\frac xy = \frac 34 \implies 4x = 3y$ so $4(n+3)=3(n+6)$ so $4n + 12 = 3n + 18$ so $4n -3n = 18-12$ so $n = 6$ is a much better, and more recommended way to do it.
H: Prove a function is continuous (using partial derivative)? Given: $\|f(x,y)_x'\|<M, \|f(x,y)_y'\|<N near (0,0)$-partial derivative- How may I prove that $f(x,y)$ is continuous in (0,0)? I tried to prove that using $\epsilon$ but got to the end of the road since partial Derivative are not related. Any help? AI: Note that $\|f(x,y)-f(0,0)\| = \|f(x,y)-f(x,0)+f(x,0)-f(0,0)\| \le \|f(x,y)-f(x,0)\| + \|f(x,0)-f(0,0)\|$. Now use the mean value theorem to bound $\|f(x,y)-f(x,0)\| $ and $ \|f(x,0)-f(0,0)\| $ separately. $\|f(x,y)-f(x,0)\| \le N \|x\|$ and $\|f(x,0)-f(0,0)\| < M\|y\|$ so $\|f(x,y)-f(0,0)\| \le M\|x\|+N\|y\| \le \sqrt{N^2+M^2}\sqrt{x^2+y^2}$. Now choose $\delta = {\epsilon \over \sqrt{N^2+M^2}}$.
H: A dense subset, $E$, of $[0,1]$ with measure $\frac{1}{2}$, and no proper subset of $E$ is an interval? I'm trying to find a subset of the unit interval that's analogous to the irrationals in some sense; it's dense in $[0,1]$, no subset of it is an interval, but it has a strictly smaller measure than the irrationals, while still having positive measure. Having measure equal to $\frac{1}{2}$ is not strictly necessary; really any $\alpha \in (0,1)$ will suffice. I'd be especially interested in seeing ig this set can be constructed, or if we can only know of it's existence via AoC. edit: After a few replies, I realize that I'm also looking for something that has a certain "uniformity" of measure. in other words, if you give me some interval, $I$, of length $\epsilon$, then $m(E \cap I)$ is the same, regardless of where $I$ is centered (as long as $I$ is fully contained in $[0,1]$, of course). So in the case of $\alpha = \frac{1}{2}$, we might be doing something like "taking the first half of the irrationals, and spreading them out evenly over all of $[0,1]$". AI: You cannot get the kind of regularity that you want: see the first result in this question.
H: Associative Law for infinitely many sets I know the associative law for (union or intersection) of two sets and I know why it works. Applying this rule many times, will show us intuitively that this rule also holds for infinitely many sets. Although, what would be an adequate proof for this generalization? Is it sufficient to say that we know that it works for two sets and it works for all the pairs, so it works for all sets? Sorry for the confusion. Intuitively seems right, but I want to understand it formally. Thank you! AI: If $\mathscr{A}$ is any family of sets, we can define $$\bigcup\mathscr{A}=\{x:\exists A\in\mathscr{A}\,(x\in A)\}$$ and $$\bigcap\mathscr{A}=\{x:\forall A\in\mathscr{A}\,(x\in A)\}\;.$$ These definitions specialize to the usual definition of $A\cup B$ and $A\cap B$ when $\mathscr{A}$ contains just two sets, and these definitions are independent of any parenthesizing or other organization of the members of $\mathscr{A}$ into smaller subcollections. Thus, the order and grouping of the sets makes no difference to the union and intersection of the family.
H: Confusion between power series and taylor series Given some power series $f(x) = \sum_{n=1}^{\infty} a_n x^n$, is it generally true that $a_n = \frac{f^{(n)}}{n!}$? If so, why? We get this form when we develop a taylor series, but why is this case for every power series? Can't the $a_n$ terms be arbitrary? I'm a bit confused. AI: Since$$f(x)=a_0+a_1x+a_2x^2+\cdots,$$you have $f(0)=a_0$ and $$f'(x)=a_1+2a_2x+3a_3x^2+\cdots\tag1$$and therefore $f'(0)=a_1$. On the other hand, since you have $(1)$, you have$$f''(x)=2a_2+6a_3x+12a_4x^2+\cdots$$and, in particular, $f''(0)=2a_2$ and so on…
H: Solution to an infinite series Given an infinite series in the form of: $$a \cdot a^{2\log(x)} \cdot a^{4\log^2(x)} \cdot a^{8\log^3(x)} \dotsb = \frac{1}{a^7} $$ find the solution for all positive and real $a$ other than $1$. The textbook, from which this problem was taken, says that there are no solutions to this equation. Given that I didn't know that before I attempted to solve it, I managed to scramble a solution which gives $x = 10^{4/7}$ and here's how I did it: First, let's take the $\log_a$ from both sides and expand by the log multiplication rule. We then obtain: $$ 1+2\log(x)+4\log^2(x)+8\log^3(x)+\dotsb = -7. $$ Let's now call the left hand side $S(x)$, so we have $S(x) = -7$. If we then go back to our last equation and subtract one from both sides and multiply by $\frac{1}{2\log(x)}$ we get: $$ 1+2\log(x)+4\log^2(x)+8\log^3(x)+\dotsb = \frac{-4}{\log(x)}. $$ So we have $S(x)$ again on the left hand side, therefore we can set the two expressions equal to one another. After a quick algebraic rearrangement, we get: $$ \log(x) = \frac{4}{7} \implies x = 10^{\frac{4}{7}} $$ I tend to believe the textbook rather than myself but don't know which of the steps was invalid. The method of substitution by $S(x)$ resembles a technique Euler supposedly used to solve some infinite fraction problems so it perhaps doesn't apply in this case with functions. AI: Your approach to find $x$ was well and good. However before assuming we have found the ultimate solution to any problem, it is good practice to substitute your solution back in the original question and see if everything checks out. That being said, notice that: $$a^{2\log x}=a^{2\times\frac47}=a^{\frac87}$$ $$a^{4\log^2x}=a^{4\times\frac{16}{49}}=a^{64\over 49}$$ Focus on the powers since we have anyway tackled the problem with $a$ by taking the logarithm. Now you can see that the powers go on increasing and each power is greater than $1$. This creates a divergent geometric series and hence can never converge to the "supposed" value of $-7$ This means that we were wrong in considering the solution to be $x=10^{\frac47}$. Since this is the only solution we could come up with, there should be no solutions for the value of $x$. And since there is no $x$ that satisfies the original equation, we cannot conclude on a value of $a$. Hence, no $a$ under the given constrains, can satisfy the given equation. $\square$
H: A simple group of order 168 doesn't have subgroups of order 14 Let $G$ be a simple group of order 168. Prove that $G$ doesn't have subgroups of order 14. I know there are eight 7-Sylows in $G$ and that the normalizer of a 7-Sylow has order 21, it was a previous exercise. My idea: Assume there is a subgroup $H<G$ of order 14 and then find a nontrivial normal subgroup. But I don't know how to proceed. Can someone give me any hints? AI: Let $H \lt G$ with $\|H\|=14$ and let $P \in Syl_7(H)$, then $P$, having index $2$, is normal in $H$, hence $H \subseteq N_G(P)$. Observe that $P \in Syl_7(G)$. But $\|G:N_G(P)\|=8$, hence $\|N_G(P)\|=21$ implying $14 \mid 21$, a contradiction to Lagrange's Theorem.
H: Does this series over $SL_2(\mathbb{Z})$ converge? Consider the following series of positive numbers $$ \sum_{\substack{a,b,c,d \in \mathbb{Z} \\ ad-bc=1}}\frac{1}{((a-b)^2 + (c-d)^2 + 1)^2} $$ Does it converge? How to prove it? AI: The answer to your literal question is "no", but for a reason which may clarify what question you actually intend to ask: Since $\pmatrix{a & b \cr c & d}\pmatrix{1 \cr -1}=\pmatrix{a-b\cr c-d}$, the expression in the denominator is the fourth power of the length of the latter vector, plus $1$. A key point is that isotropy/stabilizer subgroups in $SL_2(\mathbb Z)$ of non-zero vectors in $\mathbb Z^2$ are infinite. Here, $\pmatrix{1 & 0 \cr 1 & 1}\pmatrix{1 \cr -1}=\pmatrix{1\cr 0}$, and the isotropy subgroup of the latter vector is recognizable as $\pmatrix{1 & n\cr 0 & 1}$ for $n\in\mathbb Z$. So at the very least you may want your sum to be not over $SL_2(\mathbb Z)$, but over the coset space $SL_2(\mathbb Z)/H$, where $H$ is the isotropy subgroup of $\pmatrix{1 \cr -1}$ in $SL_2(\mathbb Z)$. In that case, your sum is a subsum of the sum of $4$th powers of lengths of non-zero vectors in $\mathbb Z^2$. (In particular, it is the sum over vectors $v=(x,y)$ with $x,y$ coprime, but we don't need this.) That is, $\sum_{x,y} {1\over (x^2+y^2)^2}$ with not both $x,y$ being $0$. Fooling around with polar coordinates and/or integral tests shows that this does converge. EDIT: the revised version of the question is significantly different, even if the formulaic assertion may seem similar. Reformulating it to help explain its meaning, let $\varphi(z,w)=1/(z-\overline{w})^{2k}$. With $2k=2$ this would essentially be a Bergman kernel for the upper half-plane. For $2k>2$, this will wind up to be a reproducing kernel for holomorphic cuspforms (something like this appeared in Petersson 1940). That is, $$ K(z,w) \;=\; \sum_{\gamma\in SL_2(\mathbb Z)} {1\over (cz+d)^{2k}}\varphi(\gamma z,w) $$ has the property that (up to a constant) $$ \int_{\Gamma\backslash\mathfrak H} K(z,w)\,f(w)\;(\Im w)^{2k} {\|dw\|\over (\Im w)^2} \;\;=\;\; f(z) $$ where $\Gamma=SL_2(\mathbb Z)$. Summing over integer matrices with determinants $N$, instead, gives the $N$th Hecke operator. (This was also known long ago. S. Lang's "Intro to modular forms" has an appendix by D. Zagier using this to compute traces. In the first edition of the book, there was an error due to the fragility of the convergence of the trace... corrected in a later edition.) A non-obvious, but conceptually robust and not-so-computational approach to proof of convergence is to "observe" that the given series is a subsum of the pullback of the Siegel-type holomorphic Eisenstein series on the Siegel upper half-space $\mathfrak H_2$ to the diagonally imbedded $\mathfrak H\times \mathfrak H$. (This was discussed in my paper, and H. Klingen's paper, in the 1984 Katata conference volume "Automorphic forms in several variables".) The convergence of those Eisenstein series was proven by H. Braun by 1939. Analogous reproducing kernels for holomorphic cuspforms exist for all the hermitian classical domains. Of course, none of this is strictly necessary to understand convergence, but it does explain why we might care, and so on.
H: Probability of 2 consecutive heads before a tails What's the probability of 2 consecutive heads before a tails, given that we keep flipping until we get either of these items? Is this approach correct? Notation: P2h = probability of 2 consecutive heads; P1h = probability of 1 head P2h = (1/2)*P1h P1h = (1/2) Thus, P2h = 1/4. If this is correct, then why is it that P(2 heads in a row) == P(2 heads in a row before 1 tail)? AI: \begin{array}{\|c\|c\|c\|} \hline \text{Coin 1} & \text{Coin 2} & \text{Probability}\\ \hline \color{red}{T} & - & 0.50\\ H & \color{red}{T} & 0.25\\ H & H & \color{blue}{0.25}\\ \hline \end{array} In any trial with two events, the favourable outcome is called a success and the other is called a failure. Intuitively, the probability of succeeding is exactly equal to the probability of succeeding before you fail, since you can only do one or the other.
H: Probability distribution for a function of random variables I'm very new in the Statistic Math field, so this question maybe be a bit trivial for you guys. Anyway, I'd appreciate any guidance in this matter. I was thinking about whether is possible to find the distribution for a function $f(x_i)$, with random variable $x_i$ satisfying $-1 \lt x_i \lt 1$, and finite $i=1,2,...,N$. For instance, suppose we have \begin{equation} f(x_i) = \frac{x_i}{\|x_{i}\|^3}, \end{equation} Would be possible to find analytically which distribution function the histogram for $f(x_i)$ follows, for finite N? If it's not, can I find the distribution for $N\rightarrow \infty$? I should also say that I have tried to implement this problem numerically, here is the result I'm getting https://drive.google.com/file/d/1G4ym5Afw8F9c17prd0daqpZCUuOtr1Xv/view?usp=sharing We clearly can see that this doesn't follow the normal distribution neither other common distributions at first sight. Also, the middle gap region is a feature attributed to the local minimum of $f(x_i)$, due to the constraint of the $x_i$ value. AI: It depends on what you mean by "find the distribution of." We can generally find $P(f(x_i) \le c)$ for most functions and random variables, so in that sense yes. But in general it will not be any named distribution like the normal distribution. It is very rare that functions (besides linear transforms) of random variables with well-known distributions also have well-known distributions because it is often easier to just work with the original distribution of $x_i$ for computing things like the mean or variance.
H: Proving the open disc in $\mathbb R^2$ is not closed I'm trying to get use to calculating limit points in metric spaces, but am getting stuck. Consider the following subset of $\mathbb{R}^{2}$ The set of all complex $z$ such that $\|z\| < 1$ I'm trying to show that this set is not closed. To do this, I need to show that it doesn't contain all of its limit points. This means I need to show that there is some neighbourhood containing no point that is different to $p$ (say). Let $D = \{(x,y) \in \mathbb{R}^{2}: \sqrt{x^{2} + y^{2}} < 1\}$ if I consider a neighbourhood $N_{r}(p) = \{q: d(p,q) < r\}$ I'm not really sure how to proceed from here. I know ultimately that the limit point will be the value $1$. I've only managed to do limit points of some very simple examples such as $\{\frac{1}{n}\}$ AI: You can prove that it is closed by proving that its complement is not open. Take $(1,0)\in D^\complement$. For each $r>0$, take $r'>0$ such that $r'<r$ and also that $r'<2$. Then $(1-r',0)\in D_r\bigl((1,0)\bigr)\cap D$. Therefore, $D_r\bigl((1,0)\bigr)\not\subset D^\complement$. This proves that $D^\complement$ is not an open set.
H: Group homomorphism from finite abelian group to $\mathbb{C}\setminus\{0\}$ Let $G$ be a finite abelian group, $g\in G$ be an element of order $n$ and $\xi = \exp (\frac{2\pi i}{n})$. I want to show that for each $0\leq i\leq n-1$, there are exactly $\frac{\|G\|}{n}$ group homomorphisms $G\to \mathbb{C}\setminus\{0\}$ which satisfy $g\mapsto \xi^i$. I applied the structure theorem of finite abelian group, but it doesn't seem to help me. AI: For a finite abelian group we have $$\|G\|=\|Hom(G,\Bbb{C}^)\|$$ Let $$f:Hom(G,\Bbb{C}^)\to \mu_n, \qquad f(\chi)=\chi(g)$$ It is surjective so $$Hom(G,\Bbb{C}^)/\ker(f)\cong \mu_n$$ Whence any coset has size $$\|Hom(G,\Bbb{C}^)\|/\|\mu_n\| = \|G\|/n$$
H: How to substitute for the variables in $f^{-1}$ of $f(x,y)$ By the horizontal line test I think that I can tell $f(x,y)=x^{2}+y^{2}-1$ is not supposed to have an inverse function, however the screenshot that I have added shows two inverse functions $ f^{-1}= \pm \sqrt{x-y^{2}+1} $. When this function just maps to $0$ how am I supposed to substitute that number into the inverse function? Example: with $x=2$ in the function $ f(x) =2x $ the outcome equals $4$, then $f^{-1}(x) = \dfrac{x}{2}$ and I use the y-value of 4 to retrieve the corresponding x-value. Wolframalpha.com AI: The $x$ in $\pm\sqrt{x+y^2-1}$ is misleading. In fact $f:\mathbb R^2\mapsto\mathbb R$ gives $f(x,y)=k$, so the inverse $f^{-1}:\mathbb R\mapsto\mathbb R^2$ takes $k$ and gives $(x,y)$. For each $k\ge 1$ we have $f^{-1}(k)$ is a set of $2$D-points representing a circle, not just one couple of $(x,y)$ values. What Wolfram is giving you is not $f^{-1}$ but the inverse of $g_y=f(\cdot,y)$ Since $g_y(x)=f(x,y)=k\iff x^2+y^2-1=k\iff x^2=k-y^2+1\iff x=\pm\sqrt{k-y^2+1}$ Thus $g_y^{-1}(k)=\pm\sqrt{k-y^2+1}$. $k$ being a "dummy" name, we can also say $g_y^{-1}(x)=\pm\sqrt{x-y^2+1}$, but this is not the "same" $x$ as the one in $f(x,y)$. This is why it largely misleading to use $x$, and $f^{-1}$ in this context. You could also be interested in $h_x=f(x,\cdot)$ or $h_x(y)=f(x,y)$ and $h_x^{-1}(k)=\pm\sqrt{k-x^2+1}$. In each case, this gives two inverse functions: $g_y^{-1}(k)$ takes $(y,k)$ as input and gives $\pm x$ in return. $h_x^{-1}(k)$ takes $(x,k)$ as input and gives $\pm y$ in return.
H: Simplifying Axler's proof that every operator on an odd-dimensional real vector space has an eigenvalue (linear algebra done right, 2nd ed) This is how the proof starts: From here, the author introduces another operator to get around the issue that $W$ might not be invariant under $T$ and follows up with some more algebra. But I think that $W$ is actually invariant, if we add the assumption that $T$ is invertible (if $T$ isn't invertible, then $0$ is an eigenvalue and we're done). It seemed odd to me that Axler might have missed that, so I wanted to verify my reasoning: Claim : Suppose $T$ is invertible. Then $W$ is invariant under $T$. Proof : If $T$ is invertible, then it's injective. So $T\|_U$ is injective, and therefore invertible (1). Now suppose for contradiction that there is some $w \in W, w \neq 0$ such that $Tw \notin W$. Since $V = U \oplus W$, then $U \cap W = \{0\} \ $. It follows that $\ Tw \in U$. From (1), it follows that $w\in U$. Contradiction. If I've got something wrong, I'm guessing it's either the very first or very last step; but it seems good to me, which is weird. AI: It is not necessarily the case that $W$ is invariant, since $W$ is any subspace for which $V = U \oplus W$. As an example, consider the transformation $T$ on $\Bbb C^2$ associated with the matrix $$ A = \pmatrix{1&2\\0&-1}. $$ The "$x$-axis" $U = \{(t,0): t \in \Bbb C\}$ is an invariant subspace. The "$y$-axis" $W = \{(0,t): t \in \Bbb C\}$ is complementary to $U$, but it is not an invariant subspace. For example, $(0,1) \in W$, but $T(0,1)\notin W$. This transformation is invertible, however. In fact, $T$ is its own inverse.
H: Finding derivative of integral $I(a)=\int_{0}^{1} \frac{dx}{1+ae^x} $ Given $$I(a)=\int_{0}^{1} \frac{dx}{1+ae^x} $$ How may I calculate $$\frac{\partial}{\partial a}I(a)=\int_{0}^{1} \frac{dx}{1+ae^x} \ ?$$ Is there any rule for this? AI: Notice, it's Leibniz integral rule which gives $$\frac{\partial}{\partial a}I(a)=\int_0^1\left(\frac{\partial}{\partial a}\frac{1}{1+ae^x}\ \right)dx=\int_{0}^{1} \frac{-e^x dx}{(1+ae^x)^2}$$
H: Differentiation in the geodesic problem Consider the geodesic problem on a flat plane. We have to find the extremal function $y$ such that: $$L[y] = \int_{a}^{b} \sqrt{1+(y')^2} \mathrm{d} x$$ is minimized, where $y(a) = y_1, y(b) = y_2$. Applying the Euler-Lagrange equation to $L(x,y,y')$. $$\frac{\partial L}{\partial y} - \frac{\partial}{\partial x} \left(\frac{\partial L}{\partial y'} \right) = 0$$ When simplifying the above equation, we say that $$\frac{\partial L}{\partial y} = 0$$ as $L$ is not dependent on $y$. My question is, how is that allowed? If $L$ is a function of $y'$, you will see a $\frac{d y'}{dy}$ term, which is equal to $$\frac{d y'}{dy}=\frac{\frac{dy'}{dx}}{\frac{dy}{dx}} = \frac{y''}{y'}$$ Am I incorrect in saying the above statement? What am I missing? AI: This is a notation problem. Write $L(y) = \int_a^b Q(x,y(x), y'(x)) dt $, with $Q(f,g,h) = \sqrt{1+h^2}$. Then Euler Lagrange says ${\partial Q(x,y(x),y'(x)) \over \partial g} = {d \over dx} H(x)$, where $H(x) = {\partial Q(x,y(x),y'(x)) \over \partial h} $. Note that ${\partial Q(f,g,h) \over \partial g} = 0$.
H: Show that $\sum_{n=1}^{\infty}\frac{x^{2}}{x^{2}+n^{2}}$ does not converge uniformly on $(-\infty,\infty)$. I am trying to prove that this infinite series $\sum_{n=1}^{\infty}\frac{x^{2}}{x^{2}+n^{2}}$ does not converge uniformly on $(-\infty,\infty)$. I can definitely show that this series converges uniformly on $[-R,R]$ for any $R$ using Weierstrass $M$-test, and I understand that even though this uniform convergence holds for all $R\in\mathbb{R}$, it may still be not true that the uniform convergence on $(-\infty,\infty)$. However, I have no doable idea to prove that the series does not converge uniformly on $(-\infty,\infty)$. My attempt is to show that $$\lim_{n\rightarrow\infty}\sup_{x\in(-\infty,\infty)}\Big\|\sum_{k=1}^{n}\dfrac{x^{2}}{x^{2}+k^{2}}-\sum_{k=1}^{\infty}\dfrac{x^{2}}{x^{2}+k^{2}}\Big\|=\lim_{n\rightarrow\infty}\sup_{x\in(-\infty,\infty)}\Big\|\sum_{k=n+1}^{\infty}\dfrac{x^{2}}{x^{2}+k^{2}}\Big\|\neq0.$$ The idea is then try to found a lower bound of $\frac{x^{2}}{x^{2}+k^{2}}$ for each $k$, and this lower bound is easy to be computed in the infinite sum. However, it is hard to find such a bound. I firstly separate the domain to $x\geq 0$ ad $x\leq 0$. For example, for $x\geq 0$, $$x^{2}+k^{2}=(x+k)^{2}-2xk\leq (x+k)^{2},$$ and thus $$\dfrac{x^{2}}{x^{2}+k^{2}}\geq \dfrac{x^{2}}{(x+k)^{2}}=\Big(\dfrac{x}{x+k}\Big)^{2},$$ but then I still cannot cancel $x$ both top and bottom so that I have a convergence series but the sup in $x$ will blow up. Is there any other way to approach this question? Thank you! AI: Note that if $\sum_{n=1}^\infty f_n$ converges uniformly in some domain $D$ then $f_n$ converges uniformly to $0$ in $D$. The proof is very similar to the standard proof that if a series of numbers converges then its general term tends to $0$. So it is enough to show the sequence of functions $f_n(x)=\frac{x^2}{x^2+n^2}$ doesn't converge uniformly to $0$ in $\mathbb{R}$. We have to show that there is some $\epsilon>0$ such that for all $n_0\in\mathbb{N}$ there are $n\geq n_0$ and $x\in\mathbb{R}$ such that $\|f_n(x)-0\|\geq\epsilon$. Indeed, let $\epsilon=\frac{1}{2}$. Then for each $n_0\in\mathbb{N}$ you can take $n=n_0$ and $x=n_0$. They satisfy $\|f_n(x)-0\|=\frac{n_0^2}{n_0^2+n_0^2}=\frac{1}{2}\geq\epsilon$.
H: Is it acceptable to think about dual vector space in terms of operations on different vector spaces? I'm looking into abstract algebra. It seems like with actual examples of covectors, the things we do the vector space operations on (addition and scalar multiplication) are objects like: $\begin{pmatrix}a&b&c\end{pmatrix}$ or $\begin{pmatrix}a\\b\\c\end{pmatrix}$, but when treating them like maps there is, at least implicitly, an operation of matrix multiplication and a dot product involved (respectively). I know the abstract definition of a dual vector space, but maybe I've having this confusion because I haven't been exposed to enough examples. At the risk of being pedantic, is thinking of a covector (or dual vector) as an element of a vector space and also an element of a structure in abstract algebra with an operation $\diamond$ present, say, such that $v^\diamond w=k$ and $v^$ is linear (k is some field element), common? You see, I basically am used to thinking of a lot of things in terms of Cayley tables, so it would make sense to me if the vectors from the dual space were on one axis of the Cayley table and vectors from the regular vector space were on the other, which would imply thinking about lots of covectors the way we do- involving an operation that has certain properties -yet I have not seen this anywhere. Is thinking of covectors as elements, under a certain operation, rather that a set of THE functions on a set, equivalent? I realize that this isn't standard, but does it work? Problem: I'm confused about whether covectors are objects or maps - I understand that its perfectly sensible to define them in terms of maps, but in practice it seems like there is always an operation involved. Question: I'm looking for 1) confirmation that covectors as maps and covectors as parts of a Cayley Table (when defined up to isomorphism), are indeed equivalent [i.e. for some vector $\alpha$ is $\alpha$ together with the operation $\diamond$, or just $\alpha$ by itself the real covector we are talking about] and 2) why this is not the standard view [to think of covectors with an operation]. AI: Implicitly, you seem to be working with a vector space over a field, let's call it $V$ over $k$. A covector usually means an element of the dual space $V^$, which is defined to be the set of linear maps $V\to k$. In particular, a covector is a type of linear map. The set of all such maps is $V^$, and you can easily check that it has a structure of a vector space under pointwise addition and scaling. Moreover, when $V$ is finite dimensional, $V\cong V^$. If we take $V$ to have a basis $e_1,\ldots, e_n$, then $v=\sum_{i=1}^n v^ie_i$ has column vector representation $$ \begin{bmatrix} v^1\\ \vdots\\ v^n \end{bmatrix}.$$ On the other hand, we can define a "dual" basis on $V^$ to be $\phi^1,\ldots \phi^n$ so that $$\phi^i(e_j)= \begin{cases} 1&i=j\\ 0&i\ne j. \end{cases} $$ Then it's clear that if $\phi=\sum_{j=1}^n a_j\phi^j$ and $v$ is as above, we have $\phi(v)=\sum_{j=1}^n a_jv^j\in k.$ So, we can view this as multiplication $$ \begin{bmatrix} a_1&\cdots&a_n \end{bmatrix} \begin{bmatrix} v^1\\ \vdots\\ v^n \end{bmatrix}=\sum_j a_jv^j. $$ In particular, we can regard a covector as a row vector as you wrote in the beginning of your post. The operation $\diamond$ you have defined is just the evaluation map $\phi\diamond v=\phi(v)$ like above. This isn't really an operation like in a group, but it does define a "bilinear" map $\diamond:V^*\times V\to k$ by $\diamond(\phi,v)=\phi(v)$. I don't think we should think of Cayley tables here. Hopefully this helps.
H: Fourier transform of $\frac{x}{x^2 + x + 1}$ Apparently the expression of the result changes depending on the sign of $\omega$, I'm not sure why. $$\mathcal{F}[f](\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{xe^{-i\omega x}}{x^2+x+1}dx$$ So I consider this complex function: $$\frac{ze^{-i\omega z}}{z^2+z+1}$$ It has poles at $-\frac{1}{2} \pm i\frac{\sqrt{3}}{2}$. I integrate over a half circle of radius $R$ in the upper half of the complex plane, closed by a line through the real numbers from $-R$ to $R$. Since: $$\frac{\|Re^{-i\theta}e^{-i\omega Re^{i\theta}}\|}{\|R^2e^{i2\theta} + Re^{i\theta} + 1\|} \leq \frac{Re^{\omega R \sin(\theta)}}{R^2 - R - 1}$$ Then if C is the circular part of the path: $$\|\int_{C}f(z)dz\| \leq \int_{C}\|f(z)\|dz \leq \int_{0}^{\pi} \frac{Re^{\omega R \sin(\theta)}}{R^2 - R - 1} d\theta$$ $$ \|\int_{C}f(z)dz\| \leq \frac{R}{R^2 - R - 1} \int_{0}^{\pi} e^{\omega R \sin(\theta)} d\theta \rightarrow 0 \space \space (R \rightarrow \infty) $$ So: $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{xe^{-i\omega x}}{x^2+x+1}dx = \frac{2\pi i}{\sqrt{2\pi}} Res(f(z), -\frac{1}{2} + i \frac{\sqrt{3}}{2}) $$ Where I used the residue over the pole at $\frac{1}{2} + i \frac{\sqrt{3}}{2}$ because I chose the circle in the upper plane, and that's it. Where does the sign of $\omega$ change the result? Thanks. AI: Note that for your range of integration, $e^{\omega R \sin \theta}$ only tends to zero provided $\Re(\omega) < 0$, which is implicitly the bound you're using. To solve this, for $\Re(\omega) > 0$ you instead close your contour in the lower half plane, and this way you pick up the other residue.
H: Partial derivative of $f(x,y)= \int_{x}^{y}e^{t^-2}dt$ I want to compute the partial derivative of the following function: $f(x,y)= \int_{x}^{y}e^{-{t^2}}dt$ Since $f$ is a continous function it has an antiderivative. Let $F$ be the antiderivative of f. Then: $\int_{x}^{y}e^{-t^2}dt = F(y) - F(x)$. Now computing the partial derivatives of $F(y) -F(x)$ results in: $\frac{\partial f}{\partial x} F(y)-F(x) = - e^{-x^2}$ $\frac{\partial f}{\partial y} F(y)-F(x) = e^{-y^2}$ Is that valid? AI: Recall from the Fundamental Theorem of Calculus, if we define ${f(x)=\int_{a}^{x}g(t)dx}$ then we have ${f'(x)=g(x)}$. In your case, a multivariate function is defined to be $${f(x,y)=\int_{x}^{y}e^{-t^2}dt}$$ Remember when computing the partial derivatives of ${f(x,y)}$, you always treat the other variable as a constant. So in actuality we get $${\frac{\partial f}{\partial y}(x,y)=e^{-y^2}}$$ Furthermore, we see $${\int_{x}^{y}e^{-t^2}dt=-\int_{y}^{x}e^{-t^2}dt}$$ And so ${\frac{\partial f}{\partial x}=-e^{-x^2}}$
H: Tangent plane equation in the point of intersection with $\space y$-axis I've encountered one tiny problem - I have to find tangent plane equation of the function $\space f(x,y)=y^3 - \sqrt{1-x^2y^2} \space$ in the point where it intersects with $\space y$-axis. So $\space f(0,y) = y^3... \space$ and that's not exactly one point, how am I supposed to find an equation with this? Is it OK if I fix an $\space y_0 \space$ and then write the equation? AI: Hints: Function $f(x,y)$ will have the intersection point with $y$-axis in a point $P=(0, y_0, 0)$. In your case it will be the point $P=(0, 1, 0)$. Find now the tangent plane... Figure:
H: Is the following restatement of convexity wrong? They claim equation 5 is restatement of convexity. What am I missing? $\lambda = 0, \lambda'=1$ seems wrong no? https://www.scihive.org/paper/1702.04877 AI: Take $F(x) = x$ on the reals, $p=0,q=1, \lambda = 1, \lambda' = 1$ then the above says $F(q)=1 \le F(p) = 0$.
H: Conditional Expectation Formula on discrete random Variable Using Indicator Function Conditioned on a discrete random variable, the conditional expectation is given by the formula : $$E(X\|Y=y)=\sum xp(x\|Y=y)$$ However I've found another formula in Wikipedia that given an event H: $$E(X\|H)=\frac{E(X 1_H)}{p(H)}$$ Can anyone provide the derivation from one formula to the other ? (H is the event Y=y) Thank you for help AI: $$\sum_x xP(x\|Y=y) = \sum_x \cfrac{xP(x, Y=y)}{P(Y=y)} = \cfrac{1}{P(Y=y)}\sum_xxP(x \cap Y=y)$$ $$X1_{Y=y} = \begin{cases} X, &Y=y \\ 0, &\text{else} \end{cases} \\ \therefore P(X1_{Y=y} = x) = P(X=x, Y=y) \therefore \sum_x xP(x \cap Y=y) = \mathsf E[X1_{Y=y}]$$
H: An Explicit isomorphism between the orthogonal Lie algebras $\mathfrak{so}_n$ and the Lie algebras of type $B_n$ or $D_n$. I know from many sources, including Samelson's Notes on Lie Algebras, that the Lie algebras of even size skew-symmetric matrices, $\mathfrak{so}_{2n}$, are isomorphic to the Lie algebras of type $D_n$ and the odd size ones are isomorphic to the Lie algebras of type $B_n$. I cannot find an explicit isomorphism at the basis level anywhere. To clarify, in the case of $\mathfrak{so}_{2n}$ for example, I'm hoping for an explicit isomorphism that sends the basis elements $\{e_{i,j}-e_{j,i}\mid 1\leq i<j\leq 2n\}$ for $\mathfrak{so}_{2n}$ to a $2n\times 2n$ matrix of the form $\left(\begin{matrix}A&B\\C&-A^T\end{matrix}\right)$ where $A,B,C\in\mathfrak{gl}_n$, $B=-B^T$, and $C=-C^T$, which is the form of the elements of $D_n$ given in Humphreys Introduction to Lie Algebras and Representation Theory. I would also love one for the odd size matrices to the basis for $B_n$ given in the same source. Could someone please provide these isomorphisms? AI: This works with exactly the kind of calculations that I did in step 2 of my answer to your previous question. Concretely, what we are doing is a base change from the symmetric bilinear form given by the identity matrix $$S_1=\pmatrix{1 & 0 &\dots & 0\\ 0&1& &0\\ && \ddots &\\ 0&0&&1}$$ (giving the skew-symmetric matrices) to the one with, depending on the parity of $n$, $$S_2=\begin{pmatrix} 1&0&0\\ 0&O&I_l\\ 0&I_l&O \end{pmatrix} \hspace{.5in}\text{or}\hspace{.5in} S_2=\begin{pmatrix} O&I_l\\ I_l&O \end{pmatrix} $$ (with $l=\lfloor n/2\rfloor$) or some scalar multiple thereof, giving the split forms as described by Humphreys, where the diagonal matrices form a nice CSA and the roots are visible. Note that in the other answer, $S_2$ was a bit different again, but that $S_2$ and Humphreys' $S_2$ are easily seen to have an easy base change between them which works over almost any field. Meaning that like in the other answer, this entire base change should work, i.e. you can find a matrix $P$ with $^tPS_1P=S_2$, if and only if the field over which we are working contains a square root of $-1$, which will come up in the base change matrix and in the explicit isomorphism. In particular, the described Lie algebras are indeed isomorphic e.g. over $\mathbb C$, but not over $\mathbb R$. (Cf. Two Definitions of the Special Orthogonal Lie Algebra, note my comment to the accepted answer.)
H: Evaluate $\oint_{\|z\|=1} \frac{dz}{\sqrt{6z^2-5z+1}} $ I want to evaluate $\oint_{\|z\|=1} \frac{dz}{\sqrt{6z^2-5z+1}} $ (Joseph Bak's chapter 12 example 2) According to the textbook, I need to change the contour to $R$ instead of $1$. But I couldn't find reason why $\oint_{\|z\|=1} \frac{dz}{\sqrt{6z^2-5z+1}} = \oint_{\|z\|=R} \frac{dz}{\sqrt{6z^2-5z+1}}$ holds for $R$ greater than 1. Can you please give me reason why it holds? AI: Answer has been edited: It's a common trick in complex analysis to parameterise the contour, and take the limit as the contour approaches some other shape (which in general will change the value of the integral, but under certain conditions it will not change the value of the integral). For example, it's used a lot when using complex integration methods to evaluate real valued integrals. Now, in this case, we are integrating the function ${\frac{1}{\sqrt{6z^2-5z+1}}}$. This has two singularities which lie in the unit circle, and you can use Residue Theorem to evaluate this. What the authour has done - he has decided to replace the unit circle with a circle of radius ${R>1}$. This is valid since, as I said before, our function is analytic on the region ${\{z \in \mathbb{C}\ \|\ 1\leq \|z\| < R\}}$. This is indeed a Theorem (Deformation Theorem, give that bad boy a Google ;D). Since ${R>1}$ can be arbitrarily large, he's approximated the function on the $R$ circle to just ${\frac{1}{\sqrt{6}z}}$ (I guess an intuitive way of thinking about it is that as ${R}$ get's larger and larger, the function becomes closer to being ${\frac{1}{\sqrt{6}z}}$ on the ${R}$ circle). This isn't rigorous, although I'm sure it could probably be rigoursly justified. Anyway, the integral of ${\oint_{\|z\|=R}\frac{1}{z}dz=2\pi i}$, and hence he's gotten $${\oint_{\|z\|=1}\frac{1}{\sqrt{6z^2-5z+1}}dz=\oint_{\|z\|=R}\frac{1}{\sqrt{6z^2-5z+1}}dz\rightarrow\frac{1}{\sqrt{6}}\oint_{\|z\|=R}\frac{1}{z}dz=\frac{2\pi i}{\sqrt{6}}}$$ Where the ${\rightarrow}$ is being used to mean ${R\rightarrow \infty}$
H: If $xM$ I am wondering how I would Prove for every pair of positive integers $x$ $y$ where such $x<y$ there exist a natural number $M$ such that if $n$ is a natural number and $n>M$ then $\frac{1}{n}<(y-x)$ So I did $y>x$ so then $y-x>0$ let $M=\frac{1}{y-x}+1$ and I guess you could use floor function on the $\frac{1}{y-x}$ So then if $n>M>\frac{1}{y-x}$ then $1/n<\frac{1}{y-x}$ not sure if this proof is ok. AI: You're on the right path. Just choose $M \in \mathbb{N}$ such that $M>\frac{1}{y-x}$ (existence is guaranteed by Archimedean property).
H: Is this subset a subspace of $\mathbb{R}^3$? $\{(x, y, z)\} \space$ with $\space x + y + z = 0$ Working through some problems in a textbook and I'm not very confident about checking if subsets are subspaces. I know that for a subset to be a subspace of $\space \mathbb{R}^3 \space$ it must be closed under addition and scalar multiplication but I'm not sure how to check this with examples. Any help would be appreciated! AI: With the help of these comments, I now have the answer! The subset IS a subspace of R3. To check if it is closed under scalar multiplication: If $x + y + z = 0$, then the following is true for any scalar multiple: $ax + ay + az = 0$ To check for addition: If $x + y + z = 0$ and $a + b + c= 0$ , then $(x+a)+(y+b)+(z+c)=0$ Therefore the subset is closed under scalar multiplication and addition, and is therefore a subspace of R3.
H: Symmetry proof in expectation Let $X_1, \dots, X_n$ be iid random variables with positive mean. Let $W_i=X_i/(X_1+\cdots+X_n)$, find $E(W_i)$. The proof goes like this: since $E(W_1+\cdots+W_n)=E(1)$, by symmetry and linearity of expectation, the answer is $1/n$. My question is, how does symmetry work here? I don't get why for example $E(X_1/(X_1+X_2))=E(X_2/(X_1+X_2))$. And more generally when can we use “by symmetry”? Thank you in advance. AI: The random variables $W_1,\dots,W_n$ are all equal in distribution since $X_1,\dots,X_n$ are identically distributed. Thus $E(W_1) = \dots = E(W_n)$. This is the "by symmetry" part of the argument. Let $\mu$ denote the law of $X_1$ (which is the same as the law of any of $X_2,\dots,X_n$ by assumption). The intuition for why $W_1,\dots,W_n$ are equal in expectation is that we form $W_1,\dots,W_n$ by taking $n$ random samples from $\mu$, and then picking any one of the samples as the numerator, and using the sum of the samples as the denominator. Since the numerator samples are all coming from the same distribution, we expect $E(W_i)$ to be a quantity that does not depend on the particular index $i$. You can prove for instance that $E(W_1) = E(W_2)$ as follows. By independence of $X_1,\dots,X_n$, the pushforward $(X_1,\dots,X_n)_P = \mu\otimes\dots\otimes \mu$ ($n$ times), so by Fubini's theorem and the change of variable $$ (x_1,x_2,\dots,x_n)\mapsto (x_2,x_1,\dots,x_n), $$ we have: \begin{align} E(W_1) &= E\bigg(\frac{X_1}{X_1+X_2+\dots+X_n}\bigg) \\ &= \int_{\mathbf R^{n}}\frac{x_1}{x_1+x_2+\dots+x_n}\,\mu(dx_1)\mu(dx_2)\dots\mu(dx_n)\\ &= \int_{\mathbf R^n}\frac{x_2}{x_1+x_2\dots+x_n}\,\mu(dx_2)\mu(dx_1)\dots\mu(dx_n)\\ &= E(W_2). \end{align*}
H: Find the orders of all elements in the group $D_3×Q$. So to start, I know that the order of an element in a group is the power that it needs to be raised to in order to equal the identity element - i.e., for $\mathbb{Z_{11}}$, the element 6 will have an order of 10 because 6^10 is the first time that it mods to 1. I'm not sure how to do this with a cross group, and I'm not sure what Q is, since it is not indicated that it is the rationals Edit $D_3$ is in fact the dihedral group of order 6, and $Q$ in this case is the quaternions AI: The rationals are commonly denoted by the blackboard $\mathbb{Q}$. The group $Q$ is probably a subset of the quaternions, namely: $Q=\{\pm 1, \pm i, \pm j, \pm k\}$ (8 elements total). $D_3$ is probably the dihedral group, which consists of: $D_3=\{\rho^0, \rho^1, \rho^2\}\cup \{s, \rho s, \rho^2 s\}$ (the symmetries of a triangle in the 2D plane, with $\rho^n$ $n$-rotations and $s$ a reflection). If we have a direct product of two groups, we can denote an element by a tuple: $a\in D_3$ and $b\in Q$ give $(a, b)\in D_3 \times Q$. With their group operation defined as: $(a, b), (c, d)\in D_3 \times Q$ gives $(a, b)\circ (c, d)=(ac, bd)$. The $\circ$ is sometimes dropped. If we have an element $(a, b)\in D_3\times Q$, suppose it has order $m$. Then, $(a, b)^m=(a^m, b^m)=(\rho^0, 1)$. What condition must $m$ satisfy for this to hold? Answer: $m$ must satisfy $m=\text{lcm}(\text{ord}(a), \text{ord}(b))$. I think with this in mind, you are able to check the cases of all the elements in this direct product. I hope this helps. Edit: There is actually a corollary of Lagrange's theorem, which states that the order of an element $x\in G$ must be a divisor of the order of $G$. This can be useful for checking the plausibility of a found order. However, note that the converse statement is not generally true.
H: Prove that the 2 sequences are nested intervals and give the element of the nested interval $$ a_{n+1} = \frac {2a_{n}b_{n}}{a_{n}+b_{n}}, b_{n+1}= \frac{a_{n}+b_{n}}{2}$$ where $0<a_{1}<b_{1}$ and $n \in \mathbb{N}$ and proven is that the sequence $[a_{1} b_{1}], [a_{2}, b_{2}]... $makes a nested interval and giving a number $c \in \mathbb{R}$ which is located in every interval of $[a_{n} b_{n}]$ I tried to use $b_{n+1}= \frac{a_{n}+b_{n}}{2}$ which is equal to $\frac{2b_{n+1}}{b_{n}}=a_{n} $, does it help to check the monotone of the sequences? AI: Note that $a_{n+1},b_{n+1}$ are the Harmonic and Arithmetic Mean of $a_n, b_n$ respectively. Now we can use the fact here that $AM\ge GM \ge HM$ . This gives $b_n\gt a_n,\forall n\in \mathbb{N}$ (since $ b_1\gt a_1$ given) $a_{n+1}-a_n=\frac{2a_nb_n}{a_n+b_n}-a_n$ $=\frac{a_n(b_n-a_n)}{a_n+b_n}\gt 0$ gives $a_{n+1}\gt a_n$ Again $ b_n-b_{n+1}=\frac{b_n-a_n}2\gt 0$ gives $b_{n+1}\lt b_n$ Thus clearly $[a_{n+1},b_{n+1}]\subset [a_n,b_n] \forall n\in \mathbb{N}$ and hence the intervals are nested Oh , I forgot about the point $c$.Thanks to the comment by @Andrei Still, I will prove why that is true. Firstly $\lim (b_n-a_n)=0$ thus $\lim b_n=\lim a_n$ The common point is given by the above common limit.Let it be $c$ Now $a_{n+1}b_{n+1}=a_nb_n=...=a_1b_1$ can be easily checked and so $c^2=\lim (a_nb_n)=a_1b_1$ and thus $c=\sqrt(a_1b_1)=$GM of $a_1$ and $b_1$
H: A sequence on $\{x_n\} \subset (0,1)$ such that $\lim_{n \to \infty} x_n=1$ with nice minimum distance properties. I am looking for a nice example of sequence $\{x_n\} \subset (0,1)$ such that $\lim_{n \to \infty} x_n=1$. $\|x_n-x_k\|= f(\|n-k\|)$ for some explicit function $f$. AI: Your conditions are impossible to fulfill. Suppose you had such a sequence $x_n$ and function $f$. Then for any $n$ $$f(\|n-(n-1)\|)=f(1)$$ is constant. But we also know $$ \lim_{n\to\infty}\|x_n-x_{n-1}\|=0$$ The only way we can have a constant distance between consecutive terms in a converging sequence is if the sequence is constant. But $x_n$ converges to $1$ which implies $x_n=1$, a contradiction to the fact that $x_n\in (0,1)$.
H: Is $8x^2-8y^2$ factorable or prime? Edit: From Blitzer Intermediate Algebra 6th edition, page 371 In exercises 23 - 48, factor completely, or state that the polynomial is prime Edit 2: These set of problems focus on factoring using sum and difference of cubes, although that's not explicitly stated in the instructions. Perhaps within that context $8x^2-8y^2$ is considered prime? My textbook says $8x^2-8y^2$ is prime, but couldn't $8$ be factored out and then factor using the difference of squares? $8x^2-8y^2$ $8(x^2-y^2)$ $8(x+y)(x-y)$ Since the textbook claims $8x^2-8y^2$ is prime, why couldn't the polynomial be factored using the above method? AI: Given the level of math that the book is about, the answer should just be $8x^2-8y^2$ is factorable.
H: Linear regression ML denotations Could somebody explain what does it mean this denotation: $$\min_w \|\|Xw - y\|\|^2_2$$ P.S. Linear regression is described this way in Machine Learning. AI: For a matrix $X$ choose $w$ such that $Xw$ is close to $y$ in the usual Euclidean distance sense.
H: There exist a compact not rectificable set What shown below is a reference from "Analysis on manifolds" by James R. Munkres So Munkres gives a counterexample of open and bounded set that is not rectificable and then as exercise he asks to find a closed and bounded set that is not rectificable. Unfortunately I can't find this such set so I ask to find it. Then I ask if there exist some condictions for which I can claim that a compact set is rectificable. So could someone help me, please? AI: A fat Cantor set is closed, bounded and of non-zero measure. It has empty interior so every point is a boundary point and every boundary point is in the set (since it is closed).
H: A sequence of $x_n \in (0,1)$ such that $x_n \to 1$ with explicit $\|x_n-x_k\|$. I am looking for a sequence of $x_n \in (0,1)$ such that 1.$x_n \to 1$ we have an explicit form for $\|x_n-x_k\|$ for every $n$ and $k AI: $$x_n := 1 - \frac{1}{n},$$ $$\|x_n - x_k\| = \|1 - \frac{1}{n} - 1 + \frac{1}{k}\| = \frac{\|n - k\|}{nk}.$$
H: How to show that a series of random variables, $\sum\limits_{n\ge1}X_n$, converges almost surely. Let $X_n$ be random variables such that for some $a_n\in \mathbb{R}$: \begin{align} \sum\limits_{n\ge1}\mathbb{P}(X_n\ne a_n)<\infty \quad \text{and} \quad \sum\limits_{n\ge1}a_n \ \ \text{converges} \end{align} Show that $\sum\limits_{n\ge1}X_n$ converges a.s. I feel like I don't know a whole lot about how to show that a series of random variables converges almost surely, so any help on this one or even general techniques to show almost sure convergence of a series would be greatly appreciated. AI: I did not see the answers before I wrote my comment so I will just post the fleshed out solution here: \begin{align} \sum\limits_{n\ge1}\mathbb{P}(X_n\ne a_n)<\infty &\implies \mathbb{P}(X_n\ne a_n \ \ \text{i.o})=0 \quad \text{by Borel-Cantelli}\\ &\implies \mathbb{P}(X_n=a_n \ \ \text{eventually})=1 \end{align} So $\exists$ a $N>0$ such that $X_n=a_n$ a.s. for all $n\ge N$ and thus, \begin{align} \mathbb{P}\big(\sum\limits_{n\ge1}X_n < \infty\big)&=\mathbb{P}\big(\sum\limits_{n\ge N}X_n < \infty\big)\\ &=\mathbb{P}\big(\sum\limits_{n\ge N}a_n < \infty\big)\\ &=1 \quad \text{by assumption} \end{align}

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